Sunny 1,000 Collection/SUNNY1000.iso/Files/Dos/Boardak/INETPUZ.ZIP/PICKOVER.TXT

Text File | 1995-02-05 | 163KB | 4,201 lines

Archive-name: puzzles/archive/pickover
Last-modified: 17 Aug 1993
Version: 4


==> pickover/pickover.01.p <==
~Title: Cliff Puzzle 1: Can you beat the numbers game?
~From: [e-mail]

If you respond to this puzzle, if possible please include your name,
address, affiliation, e-mail address.  If you like, tell me a little bit
about yourself.  You might also directly mail me a copy of your response
in addition to any responding you do in the newsgroup.  I will assume it
is OK to describe your answer in any article or publication I may write
in the future, with attribution to you, unless you state otherwise.
Thanks, Cliff Pickover

* * *
At a recent trip to the Ontario Science Center in Toronto, Canada I came
across an interesting puzzle.  The center is located minutes from
downtown Toronto and it's a vast playground of science with hundreds of
exhibits inviting you to touch, try, test, and titillate your curiosity.
The puzzle I saw there can be stated as follows.  In the 10 boxes below,
write a 10-digit number.  The digit in the first box indicates the total
number of zeros in the entire number.  The box marked "1" indicates the
total number of 1's in the number.  The box marked "2" indicates the
total number of 2's in the number, and so on.  For example, the "3" in
the box labeled "0" would indicate that there must be exactly three 0's
in the 10-digit number.

-------------------------------
| 0| 1| 2| 3| 4| 5| 6| 7| 8| 9|
| 3|  |  |  |  |  |  |  |  |  |
-------------------------------


Stop And Think

1. Is there a solution to this problem?  Are there many solutions to this
problem?

2. A more advanced an interesting problem is to continue to
generate a sequence in a recursive fashion such that each row becomes
the sequence for the previous.  For example, start with the usual
0 through 9 digits in row 1:

Row 1: 0 1 2 3 4 5 6 7 8 9

Assume Row 2 is your solution to the puzzle.  I've just inserted random
digits below so as not to give away the solution:


Row 1: 0 1 2 3 4 5 6 7 8 9   S(1)
Row 2: 9 3 2 3 3 1 6 7 8 9   S(2)
Row 3:                       S(3)

Row 2 is now the starting point, and your next job is to form row 3, row 4,
etc. using the same rules.  In the previous example, a digit in the
first box would indicate how many 9's there are in the next 10-digit number,
and so forth.

Contest: I am looking for the longest sequence of numbers users can come
up with using these rules.  Can you find a Row 2 or Row 3?
Is it even possible to generate a "row 2" or "row 3"?


==> pickover/pickover.01.s <==
1) 0 1 2 3 4 5 6 7 8 9
2) 6 2 1 0 0 0 1 0 0 0
3) 0 0 0 4 4 4 0 4 4 4
4) 6 6 6 0 0 0 6 0 0 0
5) 0 0 0 4 4 4 0 4 4 4
           .
           .
           .

and so on, repeating rows 3 and 4.
I don't know yet whether there are multiple solutions, but
I'll keep looking.

Mark Hayes
Goddard Space Flight Center (GSFC) / Interferometrics, Inc.
[e-mail]

GSFC Code 926.9
Greenbelt, MD 20771

-------------------------


In article <[e-mail]>, you write:
|> The puzzle I saw there can be stated as follows.  In the 10 boxes below,
|> write a 10-digit number.  The digit in the first box indicates the total
|> number of zeros in the entire number.  The box marked "1" indicates the
|> total number of 1's in the number.  The box marked "2" indicates the
|> total number of 2's in the number, and so on.  For example, the "3" in
|> the box labeled "0" would indicate that there must be exactly three 0's
|> in the 10-digit number.
|>
|> -------------------------------
|> | 0| 1| 2| 3| 4| 5| 6| 7| 8| 9|
|> | 3|  |  |  |  |  |  |  |  |  |
|> -------------------------------
|>
|>
|> Stop And Think
|>
|> 1. Is there a solution to this problem?  Are there many solutions to this
|> problem?

This is an old puzzle, but I'll solve it as if it was new because I
find your extension below to be interesting.
Since all possible digits must be "counted" once, the ten digits must
add up to 10.  Consider the first digit (= the amount of zeroes used):

9: Impossible, since all the other digits would have to be zero.
8: Also impossible, since we must mark a 1 under the 8, and the other
   digits then must be zeroes.
7: We must mark a 1 under the 7, and we have one more non-zero digit
   to assign.  We've used a 1, so we must put a non-zero digit under the 1.
   However, if we put a 1 there, it's wrong because we've used two ones,
   and if we put a two that's also wrong.  So 7 zeroes doesn't work.
6: Begin as before, putting a 1 under the 6.  Now we must mark under the 1,
   but putting a 1 is wrong, so put a 2.  Now we have one non-zero digit
   left to assign, and marking a 1 under the two works.  6210001000 works.
5: Now we take a different approach to analyze this.  If there are only
   five zeroes, then there are five non-zeroes, one of which is the 5 we
   marked under the zero.  Obviously a 1 must be marked under the 5 and
   zeroes under 6-9, so we have 5----10000, where the dashes contain one
   zero and three other numbers, which must add up to four (since all
   ten digits must add up to ten) - so we have two ones and a two.  But then
   the digits we have are described by 5310010000, which is not the set of
   digits we have, so there is no solution.  Similar proofs show that there
   cannot be 4,3,2, or 1 zero.
0: Impossible, since you would have to use a zero to indicate you didn't have
   a zero.

|> 2. A more advanced an interesting problem is to continue to
|> generate a sequence in a recursive fashion such that each row becomes
|> the sequence for the previous.  For example, start with the usual
|> 0 through 9 digits in row 1:
|>
|> Row 1: 0 1 2 3 4 5 6 7 8 9
|>
|> Assume Row 2 is your solution to the puzzle.  I've just inserted random
|> digits below so as not to give away the solution:
|>
|>
|> Row 1: 0 1 2 3 4 5 6 7 8 9   S(1)
|> Row 2: 9 3 2 3 3 1 6 7 8 9   S(2)
|> Row 3:                       S(3)
|>
|> Row 2 is now the starting point, and your next job is to form row 3, row 4,
|> etc. using the same rules.  In the previous example, a digit in the
|> first box would indicate how many 9's there are in the next 10-digit number,
|> and so forth.
|>
|> Contest: I am looking for the longest sequence of numbers users can come
|> up with using these rules.  Can you find a Row 2 or Row 3?
|> Is it even possible to generate a "row 2" or "row 3"?

Well, first off, our handy rule about all the digits adding up to ten no
longer applies.  Let's see if we can find an answer:

Row 1: 0 1 2 3 4 5 6 7 8 9
Row 2: 6 2 1 0 0 0 1 0 0 0
Row 3: ?

All the same digits must be placed under all the zeroes in row 2, or some
of them would be wrong, and this digit cannot be larger than 4 since six
non-zeroes are used under the zeroes in row 2.  So, consider the cases:

4: If we put 4's under all the zeroes, we must put zeroes everywhere else.
   0004440444 works.
3: Now we must place one non-zero digit under either the 6 or the 2, since
   there are two 1's that must stay alike.  Putting any non-zero digit under
   the 6 is wrong since there aren't any sixes, unless you put a 6 under
   the 6, which is still wrong.  Similarly no digit works under the two.
2: Now we must put a non-zero digit under the 2, since we already used 6 of
   them.  We must also have two zeroes, which can only go under the ones.
   This gives us --02220222.  However, we must put a non-zero under the 6,
   and we can't put a one, since we must have zeroes under the ones.  Any
   number greater than one is wrong, because we don't have that many 6's.
1: OK, we start with ---111-111, and one of the -'s must be a zero.  This
   zero must go under the 2 or the 6, because the ones must be alike (and
   we've already used some ones).  Suppose we put 6's under the ones, and
   don't use any more ones.  Then we need a 2 under the 6, and we need
   a one under the 2, which breaks what we did before.  So, instead put
   7's under the ones.  Now we must put a 1 and a 0 in the other two spots,
   but either arrangement is wrong.  We can't put a higher number under the
   ones because there aren't enough spaces left, so there is no solution
   with 1 zero.
0: Self-contradiction, as in the original problem.

So now we have a unique third row.  Can we make a fourth?

Row 1: 0 1 2 3 4 5 6 7 8 9
Row 2: 6 2 1 0 0 0 1 0 0 0
Row 3: 0 0 0 4 4 4 0 4 4 4

Now there can only be two different digits used in the next number.  Consider
the possibilities:

No zero is used: We need to mark this by putting zeroes under the zeroes
   Self-contradiction.
Some zeroes are used:  They can't go under the zeroes, so put zeroes under
   the fours.  Now six zeroes are used, so put 6's under the zeroes.
   6660006000 works.

The same logic used to find row four shows that row five must be 0004440444
again, and we get into an infinite cycle alternating between these two.


--
----w-w--------------Joseph  De [e-mail]----------------
   ( ^ )   Disclaimer: My opinions do not represent those of Owlnet.
   (O O)   Owlnet: George R. Brown School of Engineering Educational Network.
    v-v    (Unauthorized use is prohibited.)  (Being uwop-ap!sdn is allowed.)
           Snail mail: Rice U., 6100 S. Main, Houston TX 77005.
-------------------------

In rec.puzzles you write:

>Title: Cliff Puzzle 1: Can you beat the numbers game?
>From: [e-mail]

[...]
>1. Is there a solution to this problem?  Are there many solutions to this
>problem?

Yes.  No.


>2. A more advanced an interesting problem is to continue to
>generate a sequence in a recursive fashion such that each row becomes
>the sequence for the previous.  For example, start with the usual
>0 through 9 digits in row 1:

[...]

>Contest: I am looking for the longest sequence of numbers users can come
>up with using these rules.  Can you find a Row 2 or Row 3?
>Is it even possible to generate a "row 2" or "row 3"?

My program produces the following output:
        0123456789
        6210001000
        no solutions found

So I believe that the result for row 2 is unique and that there is no
result for row 3.

[ I am including the program at the end of this message just for your interest
]



>If you respond to this puzzle, if possible please include your name,
>address, affiliation, e-mail address.  If you like, tell me a little bit
>about yourself.  You might also directly mail me a copy of your response
>in addition to any responding you do in the newsgroup.  I will assume it
>is OK to describe your answer in any article or publication I may write
>in the future, with attribution to you, unless you state otherwise.
>Thanks, Cliff Pickover

The name, address etc should appear in my signature.  As for myself,
I'm a PhD student due to finish much too shortly who likes solving
puzzles.






                                                        Pauli

Paul Dale                       | [e-mail]
Department of Computer Science  | [phone]
University of Queensland        |
Australia, 4072                 | Did you know that there are 41 two letter
                                |     words containing the letter 'a'?

The program I used follows:
--------------------------------------8<------------------------------
#include <stdio.h>
#include <stdlib.h>

#define START(in) for(in=0;in<9;in++) {                 \
                        if(sum+in > 10)                 \
                                break;                  \
                        else                            \
                                sum = sum+in;           \
                        counts[digits[in]]++;

#define STOP(in)        counts[digits[in]]--;           \
                        sum -= in;                      \
                  }

main() {
short counts[10];
short i, sum;
short i0,i1,i2,i3,i4,i5,i6,i7,i8,i9;
static short digits[10] = { 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 };
short solns[10][100];
short solcnt=0;

        printf("0123456789\n");

again:
        for(i=0;i<10;i++) counts[i]=0;
        sum = 0;
        START(i0)
          START(i1)
            START(i2)
              START(i3)
                START(i4)
                  START(i5)
                    START(i6)
                      START(i7)
                        START(i8)
                          START(i9)
if(counts[0]==digits[i0] && counts[1]==digits[i1] && counts[2]==digits[i2] &&
                        counts[3]==digits[i3] && counts[4]==digits[i4] &&
                        counts[5]==digits[i5] && counts[6]==digits[i6] &&
                        counts[7]==digits[i7] && counts[8]==digits[i8] &&
                        counts[9]==digits[i9]) {
                printf("%d%d%d%d%d%d%d%d%d%d\n", i0,i1,i2,i3,i4,i5,
                        i6,i7,i8,i9);
                for(i=0;i<10;i++)
                        solns[0][solcnt] = i0;
                        solns[1][solcnt] = i1;
                        solns[2][solcnt] = i2;
                        solns[3][solcnt] = i3;
                        solns[4][solcnt] = i4;
                        solns[5][solcnt] = i5;
                        solns[6][solcnt] = i6;
                        solns[7][solcnt] = i7;
                        solns[8][solcnt] = i8;
                        solns[9][solcnt] = i9;
                        solcnt++;
                }
                          STOP(i9)
                        STOP(i8)
                      STOP(i7)
                    STOP(i6)
                  STOP(i5)
                STOP(i4)
              STOP(i3)
            STOP(i2)
          STOP(i1)
        STOP(i0)
        if(solcnt == 0) {
                printf("no solutions found\n");
        } else if(solcnt == 1) {
                for(i=0;i<10;i++)
                        digits[i] = solns[i][0];
                solcnt = 0;
                goto again;
        } else
                printf("multiple solutions found\n");
}
--------------------------------------8<------------------------------

-------------------------

In article <[e-mail]> you write:
>Title: Cliff Puzzle 1: Can you beat the numbers game?
>From: [e-mail]
>
>If you respond to this puzzle, if possible please include your name,
>address, affiliation, e-mail address.  If you like, tell me a little bit
>about yourself.  You might also directly mail me a copy of your response
>in addition to any responding you do in the newsgroup.  I will assume it
>is OK to describe your answer in any article or publication I may write
>in the future, with attribution to you, unless you state otherwise.
>Thanks, Cliff Pickover
>
>* * *
>At a recent trip to the Ontario Science Center in Toronto, Canada I came
>across an interesting puzzle.  The center is located minutes from
>downtown Toronto and it's a vast playground of science with hundreds of
>exhibits inviting you to touch, try, test, and titillate your curiosity.
>The puzzle I saw there can be stated as follows.  In the 10 boxes below,
>write a 10-digit number.  The digit in the first box indicates the total
>number of zeros in the entire number.  The box marked "1" indicates the
>total number of 1's in the number.  The box marked "2" indicates the
>total number of 2's in the number, and so on.  For example, the "3" in
>the box labeled "0" would indicate that there must be exactly three 0's
>in the 10-digit number.
>
>-------------------------------
>| 0| 1| 2| 3| 4| 5| 6| 7| 8| 9|
>| 3|  |  |  |  |  |  |  |  |  |
>-------------------------------
>
>
>Stop And Think
>
>1. Is there a solution to this problem?  Are there many solutions to this
>problem?

A. Since there are ten digits in the number, the sum of the digits in the botto
m
row must be 10.

B. If x appears under y there must be x appearences of y, hence x*y<10
So, the MAXIMUM that can appear under each number is:
---------------------
|0|1|2|3|4|5|6|7|8|9|
|9|9|4|3|2|1|1|1|1|1| max
---------------------

C. In fact, under the numbers 5..9 there can be AT MOST one non-zero (1) answer
since otherwise two numbers of the 5..9 veriaty would appear and violate rule A
.

D. So there must be at least 4 zeros. If there were exactly 4 zeros, then the
numbers 1..4 will all have under them non-zeros (as the zeros are used up for
the 5..9 group). There is also at least one number that is 5 or greater. Well,
there is a 5 (or more), a 4 (under zero), a 1 (under the 5..9 category) and
something above zero under the other 1..4 digits for a total above 10. This
violates rule A.

E. So there must be at least 5 zeros. So a (exactly one) number that is at
least 5 has a 1 under it. (since under zero would appear a >=5 number).

F. Under 1 there must be at least 1 since the solution has at least one 1 (the
one under a 5..9 number). However it could not be exactly 1 as then there would
be 2 (or more) 1's in the solution.

G. If there were 3 or more ones, then they must be under 2..9 . But then there
would be a 5 (or more) under zero + a 3 (or more) under one + a 1 under three
(or more) other places for a total above 10.

H. So there must be at exactly 2 ones in the solution. And hence, at least 1
under two.

We can summerize:

---------------------
|0|1|2|3|4|5|6|7|8|9|
|5|2|1|0|0|----1----| min
|6|2|2|1|1|----1----| max
---------------------
where the maximum under each digit is 10 - SUM(minimum of all others)

I. Since no 3 or 4 is now possible, those two numbers must have a zero under
them.

J. So there are six zeros. Hence:
---------------------
|0|1|2|3|4|5|6|7|8|9|
|6|2|1|0|0|0|1|0|0|0| min
|6|2|2|0|0|0|1|0|0|0| max
---------------------

K. Notice that "min" is a solution, while "max" is not. Hence, "min is the
*ONLY* solution!


My name is Dan Shoham. This is the only fact about me I care to make public.
You are free to attribute it, but provide me a note when you do so.

[e-mail]
-------------------------

>From [e-mail] (Chris Long) Tue Sep 15 06:08:45 1992
Path: igor.rutgers.edu!romulus.rutgers.edu!clong
~From: [e-mail] (Chris Long)
~Newsgroups: rec.puzzles
~Subject: Re: Puzzle 1 (SPOILER)
Message-ID: <[e-mail]>
~Date: 15 Sep 92 10:08:45 GMT
~References: <[e-mail]> <1992Sep15.052438.12478@qu
estrel.com>
Organization: Rutgers Univ., New Brunswick, N.J.
~Lines: 62

In article <[e-mail]>, Chris Cole writes:

Chris, don't forget to include my name on my solutions in the FAQ,
please.  My old article should be replaced with the following in the
FAQ, anyway:

--Cut here--
Solution prepared by Chris Long.

Unfortunately, this isn't completely new, since I believe a similar
puzzle I posted and answered are in the FAQ.  However, it *is* different
enough to be interesting.

In article <[e-mail]>, [e-mail] writes:

> Here's a small number puzzle :

>       Generate numbers such that the each digit in the number specifies
> the number of the occurences of the position of the digit ( postions starting
> with 0 from the left ). Example

>       The number 1210
...

My guess is only:

1210
21200

3211000
42101000
521001000
6210001000

No 1, 2, or 3 digit numbers are possible.  Letting x_i be the ith
digit, starting with 0, we see that (1) x_0 + ... + x_n = n+1 and
(2) 0*x_0 + ... + n*x_n = n+1, where n+1 is the number of digits.

I'll first prove that x_0 > n-3 if n>4.  Assume not, then this
implies that at least four of the x_i with i>0 are non-zero.  But
then we would have \sum_i i*x_i >= 10 by (2), impossible unless n=9,
but it isn't possible in this case (51111100000 isn't valid).

Now I'll prove that x_0 < n-1.  x_0 clearly can't equal n; assume
x_0 = n-1 ==> x_{n-1} = 1 by (2) if n>3.  Now only one of the
remaining x_i may be non-zero, and we must have that x_0 + ... + x_n
= n+1, but since x_0 + x_{n-1} = n ==> the remaining x_i = 1 ==> by
(2) that x_2 = 1.  But this can't be, since x_{n-1} = 1 ==> x_1>0.
Now assuming x_0 = n-2 we conclude that x_{n-2} = 1 by (2) if n>5
==> x_1 + ... + x_{n-3} + x_{n-1} + x_n = 2 and 1*x_1 + ... +
(n-3)*x_{n-3} + (n-1)*x_{n-1} + n*x_n = 3 ==> x_1=1 and x_2=1,
contradiction.

Case n>5:

We have that x_0 = n-3 and if n>=7 ==> x_{n-3}=1 ==> x_1=2 and
x_2=1 by (1) and (2).  For the case n=6 we see that x_{n-3}=2
leads to an easy contradiction, and we get the same result.  The
cases n=4,5 are easy enough to handle, and lead to the two solutions
above.
--
Chris Long, 265 Old York Rd., Bridgewater, NJ  08807-2618
--
Chris Long, 265 Old York Rd., Bridgewater, NJ  08807-2618
-------------------------


The number "2020" was left off my list by mistake ... sorry.

-Chris
-------------------------


> * * *
> At a recent trip to the Ontario Science Center in Toronto, Canada I came
> across an interesting puzzle.  The center is located minutes from
> downtown Toronto and it's a vast playground of science with hundreds of
> exhibits inviting you to touch, try, test, and titillate your curiosity.
> The puzzle I saw there can be stated as follows.  In the 10 boxes below,
> write a 10-digit number.  The digit in the first box indicates the total
> number of zeros in the entire number.  The box marked "1" indicates the
> total number of 1's in the number.  The box marked "2" indicates the
> total number of 2's in the number, and so on.  For example, the "3" in
> the box labeled "0" would indicate that there must be exactly three 0's
> in the 10-digit number.
>
> -------------------------------
> | 0| 1| 2| 3| 4| 5| 6| 7| 8| 9|
> | 3|  |  |  |  |  |  |  |  |  |
> -------------------------------
>
>
> Stop And Think
>
> 1. Is there a solution to this problem?  Are there many solutions to this
> problem?
>
[Second question and contest problem omitted]


Good puzzle!  I am wondering though whether the second question (which
I have not tried to solve yet) is moe amenable to computer search.
It seems to me that there should not be so many cases to consider, so
that even exhaustive search should work.

So, here is my ten minutes work on the first question.
I think there is a unique solution which is:  6210001000.
Here is the reasoning.
Let the number be (in Tex notation)
                d_0 d_1 d_2 d_3 d_4 d_5 d_6 d_7 d_8 d_9.
By definition
        d_0 + d_1 + d_2 + d_3 + d_4 + d_5 + d_6 + d_7 + d_8 + d_9 = 10.  (1)
Moreover, d_0 > 0, since d_0 = 0 contradicts itself.
Let d_0 = c for some integer 9 >= c >= 1.
If c = 9, then d_9 = 1, contradiction since d_1 should both be 0 and 1 then.
If 9 > c >= 1, we rewrite (1) removing all d_i s that are zeros
            c + d_(i_1) + d_(i_2) + ... + d_(i_(9-c)) = 10
        <=> d_(i_1) + d_(i_2) + ... + d_(i_(9-c)) = 10 -c       (2)
where all the d_(i_j) >= 1,  j=1,...,9-c                        (3)
(2) & (3) imply that the d_(i_j)s are 8-c 1s and one 2.
Since there exists ONE 2, then there exists at least one 1.
So the only digits in the number are 0, 1, 2, and c (if different than 1 and 2)
.
If c is either 1 or 2, we have 3 different digits in the number, which
implies d_1 <= 3, impossible since d_1 = 8 - c >= 6.
If c> 2, we have four different digits in the number, and in fact
d_0 = c, d_1 = 8-c, d_2 = 1, d_c = 1, which leaves us with 6 0s.  QED

I hope I did not miss any other cases.

Do you plan to post answers or comments later?
Leonidas

-------------------------------------------------------------------------------
-
Leonidas Palios                         The Geometry Center
                                        1300 South Second Str
[e-mail]                     Minneapolis, Minnesota 55454
-------------------------

-------------------------------
| 0| 1| 2| 3| 4| 5| 6| 7| 8| 9|
-------------------------------
| 6| 2| 1| 0| 0| 0| 1| 0| 0| 0|
| 0| 0| 0| 4| 4| 4| 0| 4| 4| 4|  <-
| 6| 6| 6| 0| 0| 0| 6| 0| 0| 0|    |
| 0| 0| 0| 4| 4| 4| 0| 4| 4| 4|  <-
  .
  .
  .


I must be missing something in my understanding of your rules.
I found the second row by imagining that I'd need lots of zeros
and putting nine in the 0 column, then skipping back and forth
adjusting things.  I had to put a tic in the 9 column, then
I had to put one in the 1 column, then I realized that had to
change that to a two since now there were two ones, and at the
same time another required tic in the 2 column balanced the
change of one to two in the 1 column, and then of course there
weren't nine zeros anymore, but there were still six and so by
changing the nine in the 1 column to a six, the one in the 9
column sould just migrate down to the 6 column.  But it almost
seems like cheating to use fours in the second row when there
were none in the second row to necessitate this kind of adjusting.
*shrug*  If this is right, the series is infinite, obviously.

Please let me know if I'm interpreting something wrong.

Thanks, and nice puzzle. :)

Grant Culbertson
[e-mail]
[e-mail]


==> pickover/pickover.02.p <==
~Title: Cliff Puzzle 2: Grid of the Gods
~From: [e-mail]

If you respond to this puzzle, if possible please include your name,
address, affiliation, e-mail address.  If you like, tell me a little bit
about yourself.  You might also directly mail me a copy of your response
in addition to any responding you do in the newsgroup.  I will assume it
is OK to describe your answer in any article or publication I may write
in the future, with attribution to you, unless you state otherwise.
Thanks, Cliff Pickover

* * *

Consider a grid of infinitesimal dots spaced 1 inch apart in a cube with
an edge equal in length to the diameter of the sun (4.5x10**9 feet).
For conceptual purposes, you can think of the dots as having unit
spacing, being precisely placed at 1.00000...., 2.00000....,
3.00000...., etc. Next choose one of the dots and draw a line through it
which extends from that dot to the edge of the huge cube in both
directions.

Stop And Think

1. What is the probability that your line will intersect another dot
in the fine grid of dots within the cube the size of the sun?
Would your answer be different if the cube were the size of the
solar system?

2. Could a computer program be written to simulate this process?

3. Answer the two questions above, but this time assume the line
to have some finite thickness, T.


==> pickover/pickover.02.s <==
-------------------------

In article <[e-mail]> you write:
>Title: Cliff Puzzle 2: Grid of the Gods
>From: [e-mail]
>
>If you respond to this puzzle, if possible please include your name,
>address, affiliation, e-mail address.  If you like, tell me a little bit
>about yourself.  You might also directly mail me a copy of your response
>in addition to any responding you do in the newsgroup.  I will assume it
>is OK to describe your answer in any article or publication I may write
>in the future, with attribution to you, unless you state otherwise.
>Thanks, Cliff Pickover
>
>* * *
>
>Consider a grid of infinitesimal dots spaced 1 inch apart in a cube with
>an edge equal in length to the diameter of the sun (4.5x10**9 feet).
>For conceptual purposes, you can think of the dots as having unit
>spacing, being precisely placed at 1.00000...., 2.00000....,
>3.00000...., etc. Next choose one of the dots and draw a line through it
>which extends from that dot to the edge of the huge cube in both
>directions.
>
>Stop And Think
>
>1. What is the probability that your line will intersect another dot
>in the fine grid of dots within the cube the size of the sun?
>Would your answer be different if the cube were the size of the
>solar system?

That depends on the manner the dot and the direction of the line were choosen.
If both process used uniform (or any other continous) distribution, then - of
course - the probability would be zero. If, for instance, the direction of
the line is always choosen to be parallel to one of the cube's edges, then the
probability is one.

>
>2. Could a computer program be written to simulate this process?

Not a meaningfull question. Simple minded programs could never simulate
infinitesimal points, but well thought out algorithm could express anything
that can be shown analytically.
>
>3. Answer the two questions above, but this time assume the line
>to have some finite thickness, T.
>

This is equivelent to making each dot of diameter T, and keeping the line thin.
For T> (1 inch / 4.5*10^9 ft) inches, the probability -> 1.

A simple minded computer program could simulate this.

Dan Shoham
[e-mail]
-------------------------

In article <[e-mail]> you write:
>1. What is the probability that your line will intersect another dot
>in the fine grid of dots within the cube the size of the sun?

About 50%, because I usually draw horizontal lines.

I.e., YOU DIDN'T GIVE THE DISTRIBUTION OF "lines".

cf the puzzle of "what is the probability that a randomly selected
chord of a circle is longer than the radius of that circle."  The
answer depends on how you "randomly select."
_________________________________________________________
Matt Crawford       [e-mail]      Fermilab


==> pickover/pickover.03.p <==
~Title: Cliff Puzzle 3: Too many 3's
~From: [e-mail]

If you respond to this puzzle, if possible please include your name,
address, affiliation, e-mail address.  If you like, tell me a little bit
about yourself.  You might also directly mail me a copy of your response
in addition to any responding you do in the newsgroup.  I will assume it
is OK to describe your answer in any article or publication I may write
in the future, with attribution to you, unless you state otherwise.
Thanks, Cliff Pickover

* * *

How many numbers have at least one digit -- a three?

In the first 10 numbers, 1,2,3,4,5,6,7,8,9,10 there is only one number
which contain the digit 3.  This means that 1/10 or 10% of the numbers
have the number 1 in the first 10 numbers.  In the first 100 numbers the
occurrence of numbers with at least one three seems to be growing.  In
fact there are 19 numbers:  3,13,23,33,43,53,63,73,83,93,
30,31,32,34,35,36,37,38,39.  This means that about 19% of the digits
contain the number 3 in the first 100 numbers.

We can make a table showing the percentage of numbers with
at least one 3-digit for the first N numbers.
N        %
10       1
100      19
1000     27
10000    34

The percentages rapidly increase to 100% indicating that almost all of
the numbers have a 3 in them!  In fact, a formula describing the
proportion of 3's can be written:  1-(9/10)**N.  The proportion gets
very close to 1 as N increases.

Stop And Think

1. How can it be that almost all of the numbers have a 3 in them?


==> pickover/pickover.03.s <==
-------------------------

You wrote (in article <[e-mail]>):
>Title: Cliff Puzzle 3: Too many 3's
>1. How can it be that almost all of the numbers have a 3 in them?

Because as the numbers get larger, they contain more digits,
increasing the probability that one of the digits in them might be a
3.  In fact, the probability that a 3 will _not_ appear in a very long
number is very low.

I like this puzzle.  Simple, but it made me think for a moment.
A three in every number?  Preposterous!  ;)

As for the other information you requested from responders: You have
my name and email address now, I don't give out my home address unless
it's necessary, and what sort of 'affiliation' are you seeking --
religious, business, or what?

     << Brian >>

--
_/_/_/       Brian Kendig  Macintosh Jedi           Live never to be ashamed
 _/_/   Starfleet Captain  Oracle Employee         if anything you do or say
  _/  Intrepid Adventurer  Saturn SL2 Owner    is published around the world
      [e-mail]  Wizard of Frobozz    -- even if what is published
    Princeton '92! BSE/CS  Writer/Actor/Singer                   is not true.
-------------------------

In article <[e-mail]> you write:
>Title: Cliff Puzzle 3: Too many 3's
>From: [e-mail]
>
>If you respond to this puzzle, if possible please include your name,
>address, affiliation, e-mail address.  If you like, tell me a little bit
>about yourself.  You might also directly mail me a copy of your response
>in addition to any responding you do in the newsgroup.  I will assume it
>is OK to describe your answer in any article or publication I may write
>in the future, with attribution to you, unless you state otherwise.
>Thanks, Cliff Pickover
>
>* * *
>
>How many numbers have at least one digit -- a three?
>
>In the first 10 numbers, 1,2,3,4,5,6,7,8,9,10 there is only one number
>which contains the digit 3.  This means that 1/10 or 10% of the numbers
>have the number 1 in the first 10 numbers.  In the first 100 numbers the
>occurrence of numbers with at least one three seems to be growing.  In
>fact there are 19 numbers:  3,13,23,33,43,53,63,73,83,93,
>30,31,32,34,35,36,37,38,39.  This means that about 19% of the digits
>contain the number 3 in the first 100 numbers.
>
>We can make a table showing the percentage of numbers with
>at least one 3-digit for the first N numbers.
>N        %
>10       1
>100      19
>1000     27
>10000    34
>
>The percentages rapidly increase to 100% indicating that almost all of
>the numbers have a 3 in them!  In fact, a formula describing the
>proportion of 3's can be written:  1-(9/10)**N.  The proportion gets
>very close to 1 as N increases.
>
>Stop And Think
>
>1. How can it be that almost all of the numbers have a 3 in them?
>

Thaddeus Crews, 509 Windsor Green Blvd., Goodlettsville, TN, 37072
Graduate Student (Ph.D.) @ Vanderbilt University, Computer Science
[e-mail]

This problem seems a little bit simple to me, but I was never that
great at math problems so I am not betting the farm on this answer.

The percentages you show for # of the first N numbers with at least
one 3-digit is also true (about) for the # of the first N numbers
with at least one 4-digit, at least one 5-digit, etc...

Basically, as N increases, so does the number of digits in N, and
therefore so does the number of chances for the digit 3 to appear
(as well as all other digits).  Given a number N with enough (?)
digits, there is a 100% chance of all digits 0-9 appearing in that
number (of course, 1.0E10000000000) does not have a 3 in it, but
if you take the next 1.0E10000000000 numbers the percent that has
a 3 will be (I suspect) 100%.

My proof is clearly weak, but the claim is this: as N increases,
the number of digits in N also increases.  As N approaches
infinity, the number of digits in N approaches infinity (at a
slower rate, however).  As the number of digits approaches infinity,
the likelyhood of any specific digit appearing at least once
approaches 100%.

I think the real question (to be answered by someone with a better
math training) would be "At what number N does the statistical
likelyhood become 100% of at least one 3-digit appearing in the
first N numbers."

Hope this helps....
--
-- Thad Crews                         (email [e-mail])
--------------------------------------------------------------------------
"Some people have a way with words, and some people, ... oh ... *not* have
a way, I suppose..."  -- Steve Martin
-------------------------

        Heh.  As the numbers get larger, they have more digits.  Assuming a ran
dom occu
various digits in the larger numbers (not unreasonable when n-> infinity) the p
r
number NOT having a 3 is very low.

        -john 'I know it's not a proof...' karakash-
-------------------------

>Title: Cliff Puzzle 3: Too many 3's

Seth Breidbart
PO Box 5157
New York, NY 10185

Morgan Stanley & Co.

>1. How can it be that almost all of the numbers have a 3 in them?

The probability that a random sequence of n digits does not contain a
3 is .9^n; as n->infinity, this probability -> 0.  Since almost all
numbers have a lot of digits (there are only a finite number of
integers with <n digits, and infinitely many with >n), the limiting
probability is 0.


-------------------------

In article <[e-mail]> you write:
>Title: Cliff Puzzle 3: Too many 3's
>From: [e-mail]
>
>If you respond to this puzzle, if possible please include your name,
>address, affiliation, e-mail address.  If you like, tell me a little bit
>about yourself.  You might also directly mail me a copy of your response
>in addition to any responding you do in the newsgroup.  I will assume it
>is OK to describe your answer in any article or publication I may write
>in the future, with attribution to you, unless you state otherwise.
>Thanks, Cliff Pickover
>
>* * *
>
>How many numbers have at least one digit -- a three?
>
>In the first 10 numbers, 1,2,3,4,5,6,7,8,9,10 there is only one number
>which contains the digit 3.  This means that 1/10 or 10% of the numbers
>have the number 1 in the first 10 numbers.  In the first 100 numbers the
>occurrence of numbers with at least one three seems to be growing.  In
>fact there are 19 numbers:  3,13,23,33,43,53,63,73,83,93,
>30,31,32,34,35,36,37,38,39.  This means that about 19% of the digits
>contain the number 3 in the first 100 numbers.
>
>We can make a table showing the percentage of numbers with
>at least one 3-digit for the first N numbers.
>N        %
>10       1
>100      19
>1000     27
>10000    34
>
>The percentages rapidly increase to 100% indicating that almost all of
>the numbers have a 3 in them!  In fact, a formula describing the
>proportion of 3's can be written:  1-(9/10)**N.  The proportion gets
>very close to 1 as N increases.
>
>Stop And Think
>
>1. How can it be that almost all of the numbers have a 3 in them?
>

No problem. In fact almost all very large numbers have all digits in them.
It is rather hard for a number with zillions of digits to avoid "3"s (or any
other digit).

It fact, the sequences "15", "172", and "666" (and any other finite sequence)
are also contained (in order) within almost all numbers.

Dan Shoham
[e-mail]

-------------------------

Before I forget:

Chris Long, 265 Old York Rd., Bridgewater, NJ  08807-2618
[e-mail]
--
Chris Long, 265 Old York Rd., Bridgewater, NJ  08807-2618
-------------------------

>Title: Cliff Puzzle 3: Too many 3's
>From: [e-mail]

>If you respond to this puzzle, if possible please include your name,
>address, affiliation, e-mail address.  If you like, tell me a little bit
>about yourself.  You might also directly mail me a copy of your response
>in addition to any responding you do in the newsgroup.  I will assume it
>is OK to describe your answer in any article or publication I may write
>in the future, with attribution to you, unless you state otherwise.
>Thanks, Cliff Pickover

>* * *

>How many numbers have at least one digit -- a three?

>In the first 10 numbers, 1,2,3,4,5,6,7,8,9,10 there is only one number
>which contains the digit 3.  This means that 1/10 or 10% of the numbers
>have the number 1 in the first 10 numbers.  In the first 100 numbers the
>occurrence of numbers with at least one three seems to be growing.  In
>fact there are 19 numbers:  3,13,23,33,43,53,63,73,83,93,
>30,31,32,34,35,36,37,38,39.  This means that about 19% of the digits
>contain the number 3 in the first 100 numbers.
>
>We can make a table showing the percentage of numbers with
>at least one 3-digit for the first N numbers.
>N        %
>10       1
>100      19
>1000     27
>10000    34

>The percentages rapidly increase to 100% indicating that almost all of
>the numbers have a 3 in them!  In fact, a formula describing the
>proportion of 3's can be written:  1-(9/10)**N.  The proportion gets
>very close to 1 as N increases.

>Stop And Think

>1. How can it be that almost all of the numbers have a 3 in them?



        I'm not sure this is the answer you are looking for, but:


        9               =       9
        9*9             =      81
        9*9*9           =     729
^S        9*9*9*9         =    6561
                etc.

The probability of having 3 as the digit in a one-digit number is 1/10.
        "       of not having 3                 "              is 9/10.

In a two-digit number, the prob. of NOT having 3 as the first digit or
the second digit, ie. not having 3 in the two-digit number, is simply
the product of (NOT having 3 in first digit) times (NOT having 3 in second):
                (9/10)*(9/10) = 81/100
                              = 0.81

For a three-digit number:  (9/10)*(9/10)*(9/10) = 729/1000
                                                = 0.729
^Q
For an n-digit number:  (9/10)**n = probability.

        We can see that as "n" becomes larger and larger, the
        probability of NOT having a three at all in the number
        becomes smaller and smaller.  Indeed, as "n" approaches
        infinity, this probability approaches zero.  In other
        words, it is very rare for a large number NOT to have 3
        as one of its digits. In fact, it is very rare for a
        large number NOT to have any of the ten possible integers
        represented at least once.



[Aside,  N      %
        10      1  = (10 - 9)/1                 times 100
        100     19 = (100 - 81)/100             times 100
        1000    27 = (1000 - 729)/1000          times 100
        10000   34 = (10000 - 6561)/10000       times 100
                        etc.                                    ]


                Kumar
                [e-mail]

ps:     I'll leave it as a small exercise to tie up the loose ends.


==> pickover/pickover.04.p <==
~Title: Cliff Puzzle 4: Time in a Bottle
~From: [e-mail]

If you respond to this puzzle, if possible please include your name,
address, affiliation, e-mail address.  If you like, tell me a little bit
about yourself.  PLEASE ALSO directly mail me a copy of your response
in addition to any responding you do in the newsgroup.  I will assume it
is OK to describe your answer in any article or publication I may write
in the future, with attribution to you, unless you state otherwise.
Thanks, Cliff Pickover

* * *

Consider a chain of bottles (B) each connected to one another by a thin
tube. A marble is placed in bottle 1.
Each tube contains a one-way valve so marbles can only
go from left to right in the tubes which are symbolized with "-" marks:

1   2   3   4
B - B - B - B -


The tubes are thin so it takes
1 hour of constant random shaking to get the marble from B1 to B2.
Likewise for each bottle.

I have not fully described the bottle collection.  Each bottle
has a backward 1-way tube to bottle 1.  I've tried to diagram these
with "*" symbols.  Each time the marble enters bottle B(N) it has
a 50% probability of going back to bottle 1 via these tubes.


****<********
*           *
***<*****   *
*       *   *
* * *   *   *
1   2   3   4
B - B - B - B -

Stop And Think

1.  In how many hours will you expect to get the marble out of bottle 10
after placing the marble in bottle 1?

2. Is there a general formula for the amount of time
required to get the ball out of bottle N into bottle N+1 given
a probability P of backwards motion (given as 50% in this problem)?

3.  In how many hours will you expect to get the marble out of bottle 10
after placing the marble in bottle 1 given two backward tubes for each
bottle instead of one backward tube?

==> pickover/pickover.04.s <==
-------------------------

~Subject: Re: Cliff Puzzle 4 (SPOILER)
~Newsgroups: rec.puzzles
~References: <[e-mail]>

In article <[e-mail]>, Cliff writes:

> 1.  In how many hours will you expect to get the marble out of bottle 10
> after placing the marble in bottle 1?

The expected amount of time to go from state n-1 to n (state 11 is an
absorbing state) is

E(n-1,n) = 1 + E(1,n)/2 for 1<n<11;

also

E(n-1,n+1) = E(n-1,n) + E(n,n+1) for 1<n<11.

If n=2 then E(1,2) = 1 + E(1,2)/2 ==> E(1,2) = 2 (it should be clear
that no E is infinite for this problem).

E(2,3) = 1 + E(1,3)/2 = 1 + E(1,2)/2 + E(2,3)/2 ==> E(2,3)/2 = 2
==> E(1,3) = 6.

I claim that in general E(1,n) = 2^n - 2 and E(n-1,n) = 2^(n-1).
Assume true for n, then E(n,n+1) = 1 + E(1,n+1)/2 = 1 + E(1,n)/2 +
E(n,n+1)/2 ==> E(n,n+1)/2 = 1 + E(1,n)/2 ==> E(n,n+1) = 2 + E(1,n)
==> E(n,n+1) = 2 + 2^n - 2 = 2^n.  Furthermore E(1,n+1) = E(1,n) +
E(n,n+1) = 2^n - 2 + 2^n = 2^(n+1) - 2.  Therefore by induction the
result is established.

Now E(1,11) = E(1,10) + 1 (ball can't go back to bottle 1 after
leaving bottle 10) = 2^10 - 1.

> 2. Is there a general formula for the amount of time
> required to get the ball out of bottle N into bottle N+1 given
> a probability P of backwards motion (given as 50% in this problem)?

I'd have to check my work, but I get E(n,n+1) = q^n, where q = 1/(1-p).

> 3.  In how many hours will you expect to get the marble out of bottle 10
> after placing the marble in bottle 1 given two backward tubes for each
> bottle instead of one backward tube?

I get E(1,n) = (q^n - q)/(q-1), so E(1,11) = E(1,10) + 1 =
^S^Q(3^10 - 3)/2 + 1.
-------------------------


In regards to your fourth problem, the following comments (marked
with a ">") should be added.  I thought the answer was quite surprising!
---

The expected amount of time to go from state n-1 to n (state 11 is an
absorbing state) is

^SE(n-1,n) = 1 + E(1,n)/2 for 1<n<11

> since we expect it'll take an hour for the ball to leave bottle n-1,
> and it then has a probability of 1/2 of returning to the first bottle;

also

E(n-1,n+1) = E(n-1,n) + E(n,n+1) for 1<n<11

> since the only way of getting to state n+1 from n-1 is to first
> go from state n-1 to n, and then from n to n+1; the total expected
> time is the sum of the two.


==> pickover/pickover.05.p <==
~Title: Cliff Puzzle 5: Mystery Sequence
~From: [e-mail]

If you respond to this puzzle, if possible please send me your name,
address, affiliation, e-mail address.  If you like, tell me a little bit
about yourself so I can cite you appropriately if you provide unique
information.  PLEASE ALSO directly mail me a copy of your response in
addition to any responding you do in the newsgroup.  I will assume it is
OK to describe your answer in any article or publication I may write in
the future, with attribution to you, unless you state otherwise.
Thanks, Cliff Pickover

* * *

What is the next term in the Mystery Sequence:

22.45906, 17600.22, 0.34714E+12,


==> pickover/pickover.05.s <==
-------------------------

Some serious roundoff error going on here, but...

The sequence 22.45906, 17600.22, 0.34714E+22 is clearly meant to be:

Pi^e, (Pi^e)^Pi, ((Pi^e)^Pi)^e,

so the next term should be (((Pi^e)^pi)^e)^pi = 1.80169E36.

Actually, it looks like you were using "pi" = 3.14159 and "e" = 2.71828, possib
l
with other intermediate rounding off, so you may have been looking for somethin
g
more like 1.8011E36.

James
[e-mail]
-------------------------

In article <+M_UUYZ8!@linac.fnal.gov> you write:
>[e-mail] (cliff) writes:
>>What is the next term in the Mystery Sequence:
>>
>>22.45906, 17600.22, 0.34714E+12,
>
>I disagree about the last couple of significant digits, but otherwise
>the series is pi^e, (pi^e)^pi, ((pi^e)^pi)^e, ... and the next term
>is about 1.8017E+36.
>_________________________________________________________
>Matt Crawford       [e-mail]      Fermilab
>
>


Background:

I recognized the approximate value of the first term from figuring
^Qout (during high school, about 20 years ago) the old question of
which is larger, e^pi or pi^e.  After that it was a mater of taking
ratios of logs of the terms.

_________________________________________________________
Matt Crawford       [e-mail]      Fermilab
     BS 1978 Applied Math & Physics; PhD 1985 Physics
-------------------------

Before I go barking up a wrong tree, I notice that

     e
   Pi   = 22.45916

>22.45906, 17600.22, 0.34714E+12,
which seems suspiciously close to your first constant. Which should I assume?

   "Typo. It should read 22.45916",
   "Coincidence.",
or "No Comment -- no clues."
^S
 ???

   /Alan Paeth
-------------------------

In article <[e-mail]> you write:
>What is the next term in the Mystery Sequence:
>22.45906, 17600.22, 0.34714E+12,

As a one-time math major, I thought I recognized that telltale 22.45906 ...

The sequence continues with 1.8016851E+36

Steve
--
-- [e-mail] -- [phone]
 __     | signature designed by | One thing about kneading that pizza dough by
(_      | (and ripped off from) | hand -- it sure gets your fingernails clean!
__)teve | Stephen Wayne Miller  |         Pizzaria Friend of Danny

==> pickover/pickover.06.p <==
~Title: Cliff Puzzle 6: Star Chambers
~From: [e-mail]

If you respond to this puzzle, if possible please send me your name,
address, affiliation, e-mail address.  If you like, tell me a little bit
about yourself so I can cite you appropriately if you provide unique
information.  PLEASE ALSO directly mail me a copy of your response in
addition to any responding you do in the newsgroup.  I will assume it is
OK to describe your answer in any article or publication I may write in
the future, with attribution to you, unless you state otherwise.
Thanks, Cliff Pickover

* * *

As many of you probably know, 5-sided stars produced by drawing a
continuous line with your pencil can nest inside each other.  (One star
can sit inside the pentagon produced by the larger star.  Each of the
5 points of the small star coincide with the 5 points of the
internal pentagon of the large star.)

Start with a five sided star formed with 5 line segments, each 1 inch
long.  Continually nest stars so that the assembly of stars gets bigger
and bigger.

Questions:
1.  How many nestings N are required to make star N
have an edge-length equal to the diameter of the sun (4.5E9 feet)?

2. How many nestings N are required to make the cumulative length
of lines of all the nested stars equal to the diameter of the sun?


==> pickover/pickover.06.s <==
-------------------------

Cliff Pickover,

So here I am, waiting to see if one of my long Grobner basis
calculations is going to finish before the machine goes down.
This is a good time to read news, and I came across this trivial
problem in rec.games.puzzles.  I'm not sure why I'm responding,
perhaps the hour, or perhaps curiousity to see what will come
of this, but I could have done this the day in high school
when I learned how to compute cos(pi/5).  The ratio between
side lengths of successive pentagrams is  r = (3+sqrt(5))/2
= 1 + golden ratio = 2.618... .   The smallest  N  for which
r^N > 5.48e10 (slightly more accurate value for sun's diameter
in inches) is 26, with r^26 = 7.37e10.  The smallest  N  for which
5[r^(N+1)-1]/(r-1) > 5.48e10 is 24, with 5(r^25 - 1)/(r-1) = 8.70e10.
This seems too trivial to post, but do with this response as you like.

Bob Holt

-------------------------


   I just started reading 'rec.puzzles', so have just seen this one and
the one before (#5)...  and to be honest I'm not sure why you put this one
out, since it is pretty straightforward.

>Start with a five sided star formed with 5 line segments, each 1 inch
>long.  Continually nest stars so that the assembly of stars gets bigger
>and bigger.

   The analytical (and general) answer to this problem comes from the
basic relationship of a "chord" of a regular pentagon, which is defined
as follows:

               _=*=_
            _=/ /   \=_
         _=/   |       \=_
      _=/      |          \=_
     *        /              *
     |       |  <-- "chord"  |
      \      |              /
       |    /              |
        \  |              /
         | /             |
          *-------------*

compared to the length of one of the sides is the golden ratio, i.e.
                  _
            1 + \/5
           ---------  .
               2

   It can then be derived that the length of the "chord" (i.e. segment
length) of the next bigger Star compared to the length of the "chord"
of its incribed Star is the square of the golden ratio, or the golden
ratio plus one, same thing.

>Questions:

>1.  How many nestings N are required to make star N
>have an edge-length equal to the diameter of the sun (4.5E9 feet)?

   Back-of-envelope calculations as follows:
^Q^S
        4.5E9 * 12 = total of 5.22E10 inches.

        ratio of Star sizes approx. 2.618.

   The best answer is 27 nested Stars, although it produces a Star with
a "chord" length of 7.366E10 inches, i.e. a bit bigger.  The first, and
smallest Star, is assumed to be the one with "chord" length of 1 inch.

>2. How many nestings N are required to make the cumulative length
>of lines of all the nested stars equal to the diameter of the sun?

   This is just the sum of a geometric sequence with the ratio being
the golden ratio squared (or the golden ratio plus one).
                                          _
                                  / 1 + \/5 \ 2
   So, S = 1 inch, and S = S     | --------- |
        0               n   n-1   \    2    /

   The sum is just the standard geometric summation, which I can't remember
offhand, but the contributing terms in the sum (other than the last term),
are less than one 1.6th of the total (by conincidence the inverse of the
golden ratio ;-).  This means that the 25th Star (term) is the determining
factor, and in this case is the answer with a total length of 8.694E10
inches amoung all of them, and 5.373E10 inches for just the sum of the
segments of the 25th Star (again, counting the first one as side length
of 1 inch, or sum of 5 inches).

   Well, that's it, I guess.  Sorry to be so exhaustive, but I like to
use analytical methods to be sure I have the right answer.

   My .signature explains most of what you need to know.  What I mean
by "Honorary Grad Student" is that I have been taking Grad math classes
since my sophomore year, and for all intensive purposes might as well
be one.  My Snail-mail address is 1521 S.W. 66th Ave., Portland, OR
97225.  As to info about myself...  I love learning about things, and
mathematics and consequently computers tend to be a great focus.

   Anyway, if you have any more puzzles, pass them along...  I am still
pondering on that sequence (puzzle #5) that you posted.

   Thanks for your time.

   Erich
--
             "I haven't lost my mind; I know exactly where it is."
   / --  Erich Stefan Boleyn  -- \        --=> *Mad Genius wanna-be* <=--
  { Honorary Grad. Student (Math) } Internet E-mail: <[e-mail]>
   \  Portland State University  /       WARNING: INTERESTED AND EXCITABLE

==> pickover/pickover.07.p <==
~Title: Cliff Puzzle 7: 3x3 Recursion
~From: [e-mail]

If you respond to this puzzle, if possible please send me your name,
address, affiliation, e-mail address, so I can properly credit you if
you provide unique information.  PLEASE ALSO directly mail me a copy of
your response in addition to any responding you do in the newsgroup.  I
will assume it is OK to describe your answer in any article or
publication I may write in the future, with attribution to you, unless
^Q^Syou state otherwise.  Thanks, Cliff Pickover

* * *

Consider the 3x3 array below.  All nine digits are used exactly once.

1 9 2
3 8 4
5 7 6

Notice that "384" is twice the number in the first row, and that
"576" is three times the number in the first row.

Questions:
1.  Are there other ways of arranging the number to produce the same
result using each digit only once and the same rules?
Remember, the second row must be twice the first.  The third row
must be 3 times the first row.

2.  Start with the number in the last row (e.g "576" or any other
solution you may find) and continue to form another 3x3 matrix using the
same rules with the new starting number.  In other words, the number in
the second row must be twice the first.  The third row must be three
times the first.  (For this problem you may truncate any digits in the
beginning.  For example, 1384 would become 384.)

Keep going.  How many matrices can you create before it is impossible
to continue.  Again, each digit must be used only once
in each matrix.

==> pickover/pickover.07.s <==
-------------------------

> Title: Cliff Puzzle 7: 3x3 Recursion
> Consider the 3x3 array below.  All nine digits are used exactly once.
> 1 9 2
> 3 8 4
> 5 7 6
> Questions:
> 1.  Are there other ways of arranging the numbers to produce the same
^Q^S^Q^S^Q^S^Q^S> result using each digit only once and the same rules?

YES .

 2 1 9       2 7 3        3 2 7
 4 3 8       5 4 6        6 5 4
 6 5 7       8 1 9        9 8 1

> same rules with the new starting number.  In other words,
> the last number becomes the first, and the number in
> the new second row must be twice the first.  The third row must be three
> times the first.  (For this problem you may truncate any digits in the
> beginning.  For example, 1384 would become 384.)
NONE. There is no solution to the continuation problem.


Bye.

Amit Agarwal
> to continue?  Again, each digit must be used only once
> in each matrix.
>
>

-------------------------

By exhaustive search I found that there are only four such arrays.
Here they are:

  1 9 2        2 1 9        2 7 3        3 2 7
  3 8 4        4 3 8        5 4 6        6 5 4
  5 7 6        6 5 7        8 1 9        9 8 1

Since these are the only four it is clear from inspection that
none of the last numbers ever begin another array with the desired
properties.

Bob Murphy ([e-mail])

-------------------------



Well, I think I have an answer to both parts.  I did what should have been
a complete analysis of all possible column combinations, but it is
certainly possible that I missed a point somewhere.  If you don't get any
answers contradicting this one, I'd be happy to send you my analysis.
Anyway - for part 1, I found the following three matrices (in additionn
to the one you gave):
                        2 1 9    2 7 3   3 2 7
                        4 3 8    5 4 6   6 5 4
                        6 5 7    8 1 9   9 8 1

Note that the first one of these can be generated from yours by moving
the third column to the first position, and the third one can be generated
from the second similarly.  In both cases, one column does not receive
or produce any carryovers, so it can be placed at either end.

For part 2, there is obviously no solution, assuming that these are indeed
the only four matrices satisfying the requirements.  In my analysis, I
included columns with carryovers in all positions, so if there were any
^Qmatrices that would satisfy the relaxed condition of part 2 I should
have found them.

                                                        Dan Blum
                                                        Institute for the
                                                        Learning Sciences
                                                        Northwestern U.
                                                        [e-mail]
-------------------------


Cliff,

In article <[e-mail]> (Bob Murphy) writes:

>By exhaustive search I found that there are only 4 starting numbers
>which produce a 3x3 array with the desired property.  Here they are:
>
>   1 9 2        2 1 9        2 7 3        3 2 7
>   3 8 4        4 3 8        5 4 6        6 5 4
>   5 7 6        6 5 7        8 1 9        9 8 1


For each of these solutions I happened to notice that the sum of each row
is a constant:

        sum(row1) = 12
        sum(row2) = 15
        sum(row3) = 18

(necessary but not sufficient condition)

And the sums all differ by the same constant (3).  I wonder if this
property may somehow be generalized to matrices of higher degree?


Regards,


-- Greg Schmidt         [e-mail]

-------------------------


> If you respond to this puzzle, if possible please send me your name, address,
> affiliation, and e-mail address so I can credit you in the future if needed.
> If you like, tell me a little bit about yourself so I can cite you
> appropriately if you provide unique information.  PLEASE ALSO directly mail
> me a copy of your response in addition to any responding you do in the
> newsgroup.  I will assume it is OK to describe your answer in any article or
> publication I may write in the future, with attribution to you, unless you
> state otherwise.
> Thanks, Cliff Pickover
>
> Consider the 3x3 array below.  All nine digits are used exactly once.
>
> 1 9 2
> 3 8 4
> 5 7 6
>
> Notice that "384" is twice the number in the first row, and that
> "576" is three times the number in the first row.
>
> Questions:
> 1.  Are there other ways of arranging the numbers to produce the same
> result using each digit only once and the same rules?
> Remember, the second row must be twice the first.  The third row
> must be 3 times the first row.
>
> 2.  Start with the number in the last row (e.g "576" or any other
> solution you may find) and continue to form another 3x3 matrix using the
> same rules with the new starting number.  In other words,
> the last number becomes the first, and the number in
> the new second row must be twice the first.  The third row must be three
> times the first.  (For this problem you may truncate any digits in the
> beginning.  For example, 1384 would become 384.)
>
> Keep going.  How many matrices can you create before it is impossible
> to continue?  Again, each digit must be used only once
> in each matrix.

Well, this is probably not news to you by now, but I only get four solutions
to the original problem:

1 9 2      2 1 9      2 7 3      3 2 7
3 8 4      4 3 8      5 4 6      6 5 4
5 7 6      6 5 7      8 1 9      9 8 1

If we relax the rules slightly and allow zeroes, and just specify that the nine
numbers only have to be different, then we get two more solutions:

0 7 8      2 6 7
1 5 6      5 3 4
2 3 4      8 0 1

both of which use the digits 0-8, which may be of interest.

The second problem (in either form) has only the above solutions, with only one
^S^Qmatrix in each solution.

If we switch to base 9 (where we must use a zero), there is no solution to the
first, and only one solution to the second (with only one matrix):

4 8 1
0 7 2
5 6 3

In fact, I considered 3 versions of problem 2.  In all cases zeroes were
allowed, but the 9 numbers had to be different.  For each of them the first 3x3
matrix has to meet the original specifications; where they differ is in how the
succeeding matrices are constructed.  In the ensuing discussion, the original
number is called 'n'.  So in the example given with the problem, n is 192.

Version A:  The second matrix consists of rows with n*2, n*3 and n*4 in them.
           (The last three digits of those, anyway.)  The next would have n*3,
           n*4, and n*5, then n*4, n*5, n*6, etc.

Version B:  The second matrix consists of n*3, n*6, n*9.  (This is essentially
            the second problem as given.)

Version C:  The second matrix consists of n*4, n*5, n*6.  The next would have
            n*7, n*8, n*9 etc.

Results for various bases:

Base 9:
  A, B, C:   4 8 1
             0 7 2
             5 6 3

Base 10:
  A, B, C:   0 7 8    1 9 2    2 1 9    2 6 7    2 7 3    3 2 7
             1 5 6    3 8 4    4 3 8    5 3 4    5 4 6    6 5 4
             2 3 4    5 7 6    6 5 7    8 0 1    8 1 9    9 8 1

  In addition, with version C, we get a second matrix for 219.

     2 1 9     8 7 6
     4 3 8 ==> 0 9 5
     6 5 7     3 1 4

Base 11:   (A, B, etc. represent 10, 11, etc..)
  A, B, C: 18 solutions.  From now on, I'll only show the multiple matrix ones.

  A:  5 A 1     0 9 2    6 7 4     2 3 8
      0 9 2 ==> 6 8 3    2 3 8 ==> 9 0 1
      6 8 3     1 7 4    9 0 1     4 7 5

  B:  9 3 4     5 A 1
      7 6 8 ==> 0 9 2
      5 A 1     6 8 3

  C:  8 9 1     2 3 4
      6 7 2 ==> 0 1 5
      4 5 3     8 A 6

  (Note that the B solution ends in an A solution matrix!)

Base 12:   19 solutions

  A:  7 3 4     2 6 8
      2 6 8 ==> 9 A 0
      9 A 0     5 1 4

  B:  None

  C:  3 5 7     1 A 4
      6 B 2 ==> 5 3 B
      A 4 9     8 9 6

Base 13:   71 solutions...and it rapidly increases from here....

The number of solutions rises rapidly, as we might expect, as the more possible
values for digits there are in the base, the more likely the set of 9 will be
distinct.  If we look at solutions which only involve the digits 1-9, then the
following is a list of all solutions (for all bases):

Base 10:  1 9 2    2 1 9    2 7 3    3 2 7
          3 8 4    4 3 8    5 4 6    6 5 4
          5 7 6    6 5 7    8 1 9    9 8 1

Base 11:  7 8 3    8 4 6    8 9 1
          4 5 6    5 9 1    6 7 2
          1 2 9    3 2 7    4 5 3

(Tested all cases until base 17.  After that, no solution can involve a carry.
But there are no solutions without carries.  So, no more solutions.)

I hope this is of some interest.

Cheers,
Geoff.

-------------------------------------------------------------------------------
Geoff Bailey (Fred the Wonder Worm)   |   Programmer by trade --
[e-mail]                      |       Gameplayer by vocation.
-------------------------------------------------------------------------------


-------------------------


> Ref:  Your note of Mon, 19 Oct 92 22:24:47 EST
>
> Where are you from?

Whoops, knew I forgot to put something in.  I'm currently a student at the
University of Sydney (Australia), doing Computer Science (Honours).

Cheers,
Geoff.

-------------------------------------------------------------------------------
Geoff Bailey (Fred the Wonder Worm)   |   Programmer by trade --
[e-mail]                      |       Gameplayer by vocation.
-------------------------------------------------------------------------------


-------------------------


By the way, I tried searching for analogous solutions for other sizes and other
bases.  So the new problems become:

Consider an n by n matrix containing the 'digits' from 1 to n^2 in a base b,
where b > n^2.  The i'th row of the matrix consists of the last n 'digits' of
i*(first row).  The versions differ in how succeeding matrices may be
constructed.  Let f be the first row.

Version A:  The next matrix has rows with 2*f, 3*f, ... , (n+1)*f
           The j'th matrix has rows j*f, (j+1)*f, ... , (n+1-j)*f

Version B:  The next matrix has rows with n*f, 2*n*f, ... , n*n*f
           The j'th matrix has rows (n^(j-1))*f, 2*(n^(j-1))*f, .... , (n^j)*f

Version C:  The next matrix has rows with (n+1)*f, (n+2)*f, ... , 2*n*f
           The j'th matrix has rows (1+n*(j-1))*f, (2+n*(j-1))*f, ..., j*n*f

Basically these are analogies of the three versions I wrote to you about
before.  The results I get are:

n: 1    base: any above 1

    A, B, C:     1

n: 2    base: 5

    A, B, C:      3  2             4  1
                  1  4             3  2

    In case B, the second one extends:

                  4  1 ==> 3  2
                  3  2     1  4

    In case C, the second one also extends:

                  4  1 ==> 2  3
                  3  2     1  4

        base: 6

    A, B, C:      1  4             3  4
                  3  2             1  2

  Note that the only solution to the first problem (no overflow allowed) is
                  1  4     (in base 6)
                  3  2

n: 3    base: 10

    A, B, C:  1  9  2     2  1  9     2  7  3     3  2  7
              3  8  4     4  3  8     5  4  6     6  5  4
              5  7  6     6  5  7     8  1  9     9  8  1

        base: 11

    A, B, C:  7  8  3     8  4  6     8  9  1
              4  5  6     5  9  1     6  7  2
^S^Q^S              1  2  9     3  2  7     4  5  3

  Note the base 10 solutions all solve the first problem, while none of the
  base 11 solutions do, and there is no second matrix for any of them.

n: 4    base: 18

    A, B, C:  1 15 14  4    1 15 16  2    2  1 15 16    2  3 13 16
              3 13 10  8    3 13 14  4    4  3 13 14    4  7  9 14
              5 11  6 12    5 11 12  6    6  5 11 12    6 11  5 12
              7  9  2 16    7  9 10  8    8  7  9 10    8 15  1 10


              3 13 14  4    3 13 16  2    4  1 15 14    4  3 13 14
              7  9 10  8    7  9 14  4    8  3 13 10    8  7  9 10
             11  5  6 12   11  5 12  6   12  5 11  6   12 11  5  6
             15  1  2 16   15  1 10  8   16  7  9  2   16 15  1  2


    In case C, two of them extend:

   1 15 16  2      9  7  8 10      2  1 15 16     10  9  7  8
   3 13 14  4 ==> 11  5  6 12      4  3 13 14 ==> 12 11  5  6
   5 11 12  6     13  3  4 14      6  5 11 12     14 13  3  4
   7  9 10  8     15  1  2 16      8  7  9 10     16 15  1  2

  Note all of these solutions solve the first problem (no overflow).

Unfortunately, my algorithm is O((n!)^2), so any results for n = 5 are not
going to be forthcoming soon.

Cheers,
Geoff.

-------------------------------------------------------------------------------
Geoff Bailey (Fred the Wonder Worm)   |   Programmer by trade --
[e-mail]                      |       Gameplayer by vocation.
-------------------------------------------------------------------------------



==> pickover/pickover.08.p <==
~Title: Cliff Puzzle 8: Squares and Squares and Squares ....
~From: [e-mail]

If you respond to this puzzle, if possible please send me your name,
address, affiliation, e-mail address, so I can properly credit you if
you provide unique information.  PLEASE ALSO directly mail me a copy of
your response in addition to any responding you do in the newsgroup.  I
will assume it is OK to describe your answer in any article or
publication I may write in the future, with attribution to you, unless
you state otherwise.  Thanks, Cliff Pickover

* * *

1.  What is the smallest square with leading digit 1 which remains a
square when the leading 1 is replaced by a 2?

In other words, if x**2 = 1.........., is there a y**2 = 2.........  ?

2.  What is the smallest square with leading digit 1 which remains a
square when the leading 1 is replaced by a 2 and also remains a square
when the leading digit is replaced by a 3?

3.  What is the smallest square with leading digit 1 which remains a
square when the leading 1 is replaced by a 2, and also remains a square
^Q^S^Qwhen the leading digit is replaced by a 3, and also remains a square
when the leading digit is replaced by a 4?

==> pickover/pickover.08.s <==
-------------------------


> 1.  What is the smallest square with leading digit 1 which remains a
> square when the leading 1 is replaced by a 2?
>
    11025 ( 105 * 105 )      ----       21025 ( 145 * 145 )

>
> 2.  What is the smallest square with leading digit 1 which remains a
> square when the leading 1 is replaced by a 2 and also remains a square
> when the leading digit is replaced by a 3?
>
    No solution till 1,000,000,000.

> 3.  What is the smallest square with leading digit 1 which remains a
> square when the leading 1 is replaced by a 2, and also remains a square
> when the leading digit is replaced by a 3, and also remains a square
> when the leading digit is replaced by a 4?
>
>
   No solution till 1,000,000,000.


The property that you are looking for ( however with different leading

digits ) is owned by the following numbers.


2025    3025
-------------
11025   21025
57600   67600
---------------
202500   302500
342225   442225
------------------
1102500   2102500
3515625   4515625
5760000   6760000
-------------------
11390625   21390625
20250000   30250000
34222500   44222500
----------------------
110250000     210250000
196700625     296700625
351562500     451562500
576000000     676000000
-------------------------

This is probably of no use to you, but, anyway.

-------------------------

In article <[e-mail]> you write:
>Title: Cliff Puzzle 8: Squares and Squares and Squares ....
>1.  What is the smallest square with leading digit 1 which remains a
>square when the leading 1 is replaced by a 2?

>In other words, if x**2 = 1.........., is there a y**2 = 2.........  ?

(Isn't this first part an old puzzle?)

105^2=11025; 145^2=21025.  In general we want 10^k=(y-x)(y+x) and
1.5 < (y/x)^2 < 2.  Thus y+x and y-x must be factors of 10^k of
the same parity whose ratio is between 5.828... and 9.899...
(these are (t+1)/(t-1) for t^2=2 and 1.5 respectively).  The
smallest solution (x,y)=(105,145) corresponds to the factorization
10^4=40*250; gp/pari's "fordiv" function allows one to easily list
all primitive solutions [i.e. not obtained from a smaller solution
by multiplying x,y by the same power of 10] with x^2 and y^2 each
having at most (say) 50 digits:

[x,y]=

[145, 105]
[17225, 14025]
[454625, 326625]
[53948125, 43708125]
[1425503125, 1015903125]
[168971890625, 136203890625]
[529265958203125, 424408358203125]
[1657888279384765625, 1322343959384765625]
[5193483785077392578125, 4119741961077392578125]

In fact it can be seen that the primitive solutions correspond to
integer linear combinations of log(2) and log(5) lying in a certain
fixed interval (determined by the bounds 5.828... and 9.899...),
which probably explains the regular growth of this list.

>2.  What is the smallest square with leading digit 1 which remains a
>square when the leading 1 is replaced by a 2 and also remains a square
>when the leading digit is replaced by a 3?

There is no such beast, since the three squares would constitute an
arithmetic progression of integer squares of common difference 10^k,
and so give an A.P. of 3 rational squares of common difference 1 or 10 ---
which is known to be impossible by a "2-descent" argument (the case of
common difference 1 is already due to Fermat).  [We were lucky here:
in a different number system this argument might fail; for instance the
squares of 7/2, 17/2, 23/2 are an A.P. of common difference 60, the
sexagesimal base.  (Some numerology: 7,17,23 are the first three primes
of which 2 is a quadratic residue.)  Still, given the base b, the general
theory of elliptic curves indicates that the rational solutions of
Y^2-X^2=Z^2-X^2=b are rather sparsely distributed (the number of d-digit
solutions growing as some power of d), and the extra condition that they
arise by changing only the initial digits of three integer squares is
strong enough to ensure that there are at most finitely many solutions;
with yet more powerful methods one can even provably list them all.]

>3.  What is the smallest square with leading digit 1 which remains a
>square when the leading 1 is replaced by a 2, and also remains a square
^S^Q^S^Q^S>when the leading digit is replaced by a 3, and also remains a square
>when the leading digit is replaced by a 4?

Of course the above solution to part 2 also disposes of this part;
alternatively I could apeal to another classic result of Fermat:
there is no 4-term A.P. of rational squares.

My question: why all the blank spaces at the end of every line?

--Noam D. Elkies ([e-mail])
  Dept. of Math., Harvard Univ., Cambridge, MA 02138
-------------------------

I dunno the direct answer to your squares problem.  I do know,
however, that phi (from the Golden Ratio--approx 0.61), which is
defined as the number x such that  x + 1 = x^2.  It just so happens
that phi+1 and (phi+1)^2 differ by only 1.  (1.61 and 2.61)  The
rest of the digits are the SAME!  Phi = (Sqrt(5)-1)/2.

Phi+1 = (Sqrt(5)+1)/2    phi+2 = (Sqrt(5)+3)/2
(Phi+1)^2= (5+2*Sqrt(5)*1+1)/4 = (2*Sqrt(x)+6)/4 = (Sqrt(x) + 3)/2

Notice how that all works out?  Perhaps this property will bring you
closer to an answer.  I just sent you all my personal data in
a previous letter concerning your 123 problem.  Let me know
what you think of this approach, ok?  Thanks in advance!

--Joseph Zbiciak  [e-mail]


-------------------------

In article <[e-mail]> you write:
: 2.  What is the smallest square with leading digit 1 which remains a
: square when the leading 1 is replaced by a 2 and also remains a square
: when the leading digit is replaced by a 3?

This is not possible.  One of these numbers would leave a remainder
of 2 when divided by 3, and no square is congruent to 2 modulo 3.

--
David Radcliffe
[e-mail]
-------------------------


In article <[e-mail]> you write:
: 1.  What is the smallest square with leading digit 1 which remains a
: square when the leading 1 is replaced by a 2?

11025.  I found, by hand, all integral solutions to
(x+y)(x-y) = 10000.  The solution (145,105) is the only
one with 10000 < y^2 < 20000.

You have permission to use my solution, but not my name.

--
David Radcliffe
[e-mail]
-------------------------

Well, as a previous poster already mentioned on Rec.puzzles, there are only 4
solutions to the initial problem. They are 192, 219, 293, and 327. None of
these solutions can be connected to others as in part 2 of your problem.

I first extended the problem to allow any multipliers. So the second row must
be some multiple of the first and the third some other multiple of the first.
I found 19 solutions to this problem. However, there is still no way to chain
a second solution to the first.

Then I allowed 0s. Now there are 134 solutions. There are also 17 2-chains.
There are two 3-chains which I will list here:
    192     394
*2= 384 *3=1182
*3= 576 *4=1576
*7=4032  now the same as the other solution.
*9=5184
*4= 736
*5= 920

I will be more than happy to send you all 134 solutions if you really want
them! I also have Pascal source code.

Comments on some of your other problems will follow.

Dan Cory
Senior, Stanford

perm. address:
55 Cedar St.
Chapel Hill, NC 27514

school address:
PO Box 13113
Stanford, CA 94309

Should you use any of my results, please send a copy of the work to the
^Qpermanent address above.
-------------------------


In article <[e-mail]>, you write:
|> Title: Cliff Puzzle 8: Squares and Squares and Squares ....
|> 1.  What is the smallest square with leading digit 1 which remains a
|> square when the leading 1 is replaced by a 2?

11025 = 105^2, 21025 = 145^2.

^S|> 2.  What is the smallest square with leading digit 1 which remains a
|> square when the leading 1 is replaced by a 2 and also remains a square
|> when the leading digit is replaced by a 3?
|>
|> 3.  What is the smallest square with leading digit 1 which remains a
|> square when the leading 1 is replaced by a 2, and also remains a square
|> when the leading digit is replaced by a 3, and also remains a square
|> when the leading digit is replaced by a 4?

These two cases never occur.

Proof: (This was a LOT harder than I thought it would be when I started!)
The original problem can be reduced to:
 "Find positive integers x,y,n such that
    y^2-x^2 = 10^n  and  10^n < x^2 < 2*10^n." [1]

The second problem amounts to finding x,y,z,n which meet the above
conditions, plus z^2-y^2=10^n.

For the second problem, look at the set of solutions to
  z^2-y^2 = 10^n,  2*10^n < y^2 < 3*10^n.  [2]

A solution to the second problem consists of x,y,z,n, where x,y,n solve
the original problem and y,z,n solve the above system.

The first equation in [1] can be factored into (y-x)(y+x) = 10^n = 2^n * 5^n.
Similarly (z-y)(z+y) = 10^n.  Since x,y,z are integers, we must have
y+x = 2^a * 5^b,  y-x = 2^(n-a) * 5^(n-b)
z+y = 2^c * 5^d,  z-y = 2^(n-c) * 5^(n-d)
where a,b,c,d are integers.  When a=c and b=d, y+x = z+y and y-x = z-y,
which leads to a contradiction.

Then 2y = 2^a * 5^b + 2^(n-a) * 5^(n-b) = 2^c * 5^d + 2^(n-c) * 5^(n-d)
However, in the last equality above, divide both sides by 2^f, where f is
the smallest of a, c, n-a, and n-c. The result is:

2^(a-f) * 5^b + 2^(n-a-f) * 5^(n-b) = 2^(c-f) * 5^d + 2^(n-c-f) * 5^(n-d)   [3]

Now, at least one of the four products above is a product of only 5's, and
is odd.  Only one is odd unless a=c, 2a=n, or 2c=n.
    If a=c, then either b=d (contradiction) or z+y is at least
  a factor of 5 larger than y+x.  However, considering
    sqrt(3)*sqrt(10^n) < z < 2*sqrt(10^n)
    sqrt(2)*sqrt(10^n) < y < sqrt(3)*sqrt(10^n)
            sqrt(10^n) < x < sqrt(2)*sqrt(10^n)
  we have:
    (sqrt(3)+sqrt(2))*sqrt(10^n) < z+y <       (2+sqrt(3))*sqrt(10^n)
          (1+sqrt(2))*sqrt(10^n) < y+x < (sqrt(3)+sqrt(2))*sqrt(10^n)
^Q^S^Q^S^Q  and then (z+y)/(y+x) < (2+sqrt(3))/(1+sqrt(2)) < 5.
    If a exactly equals n/2:
  In the case that b=a=n/2, y+x = y-x, so x=0 (not possible).
  If b<n/2, y-x>y+x, but we want x to be positive, so b>n/2.  Since b and
  n/2 are integers (remember n/2=a), b-(n-b) >= 2, and (y+x)/(y-x) >= 25.
  This gives     (y+x) >= 25(y-x),
        (y+x+y-x) = 2y >= 26(y-x),
                     y >= 13y-13x,
                   13x >= 12y,
                   x/y >= 12/13
               x^2/y^2 >= 144/169

  However, we know 10^n < x^2 < 2*10^n, and y^2 = x^2 + 10^n, so x^2/y^2
  varies between 1/2 and 2/3, and cannot be greater than 144/169.
    Similarly, when c=n/2, the same argument applies, and in the final step
  we know y^2/z^2 varies between 2/3 and 3/4.
Finally, we've eliminated all cases where more than one of the terms in [3]
is odd.  With exactly one term odd, we have odd=even, a contradiction,
so there is no solution.

--
----w-w--------------Joseph  De [e-mail]----------------
   ( ^ )   Disclaimer: My opinions do not represent those of Owlnet.
   (O O)   Owlnet: George R. Brown School of Engineering Educational Network.
    v-v    (Unauthorized use is prohibited.)  (Being uwop-ap!sdn is allowed.)
-------------------------


G'day Cliff!

> * * *
>
> 1.  What is the smallest square with leading digit 1 which remains a
> square when the leading 1 is replaced by a 2?
>
> In other words, if x**2 = 1.........., is there a y**2 = 2.........  ?

The smallest I could find was 105**2 = 11025
                              145**2 = 21025

Indeed, an exhaustive search shows that this is the smallest.

The other pairs I found (after a few minutes playing with pen and paper - I
could probably write a program to generate them ad nauseum, but I've got a
draft thesis to write...) were:

    3375**2 = 11390625,       4625**2 = 21390625
^S^Q^S^Q   14025**2 = 196700625,     17225**2 = 296700625
  326625**2 = 106683890625, 454625**2 = 206683890625

I don't know what pattern there is in them.  Of course, if x is a solution,
then so is 10*x.  So these give solutions for 1050*1050 = 1102500, etc.

> 2.  What is the smallest square with leading digit 1 which remains a
> square when the leading 1 is replaced by a 2 and also remains a square
> when the leading digit is replaced by a 3?
>
> 3.  What is the smallest square with leading digit 1 which remains a
> square when the leading 1 is replaced by a 2, and also remains a square
> when the leading digit is replaced by a 3, and also remains a square
> when the leading digit is replaced by a 4?

I'll answer part 3 first.  If such a square exists, then observe that we have
4 squares in arithmetic progression (common difference a power of 10).  There
is a well known theorem that there is no set of four squares in arithmetic
progression, so there is no solution to part 3.

Now, for part 2.  We have 3 squares in arithmetic progression.  Another well
known (and not too hard to derive) theorem states that for three squares in
arithmetic progression, their common difference is of the form:

D = 4 * K^2 * m * n * (m^2 - n^2) = 4 * K^2 * m * n * (m + n) * (m - n)

Now, this value is a power of 10.  So the only primes in its factorisation are
2 and 5.  Hence neither m nor n is divisible by 3.  So (m^2 - n^2) is
divisible by 3.  Hence a power of 10 is divisible by 3.  Contradiction.  So
now such set of three squares exist (which also proves part 3).

Cheers,
Geoff.

PS: I assume you still have whatever details of mine you care about.

-------------------------------------------------------------------------------
Geoff Bailey (Fred the Wonder Worm)   |   Programmer by trade --
[e-mail]                      |       Gameplayer by vocation.
^S-------------------------------------------------------------------------------


-------------------------

Here is the solution I just posted to rec.puzzles. Note that I changed my mind
on this puzzle!

Dan Cory
Senior, Stanford
PO Box 13113
Stanford, CA 94309
[e-mail]

~Newsgroups: rec.puzzles
~Subject: Re: Cliff Puzzle 8: Squares and Squares ... (SPOILER)
Approved: [e-mail]
Summary: solutions to part 1, no solutions to parts 2 or 3
Expires:
~References: <[e-mail]>
~Sender:
Followup-To:
^QDistribution:
Organization: DSG, Stanford University, CA 94305, USA
Keywords: squares, cliff, 8, gcd

In article <[e-mail]> [e-mail] (cliff)
>1.  What is the smallest square with leading digit 1 which remains a
>square when the leading 1 is replaced by a 2?
>In other words, if x**2 = 1.........., is there a y**2 = 2.........  ?

We write this condition as the following equations with x,y,a integers:
y^2-x^2=10^a
1*10^a<=x^2<=2*10^a
2*10^a<=y^2<=3*10^a
We factor the first equation:
(y-x)(y+x)=10^a.
Let u=x+y. Then 10^a/u=x-y. Since x+y>x-y, u>10^a/u so u>10^(a/2)
Then x=(u-10^a/u)/2 and y=(u+10^a/u)/2.

Subsitute these equations into the inequalities above.
For x we get:
10^a<=((u-10^a/u)/2)^2<=2*10^a
Take the square root of both (all three?) sides:
10^(a/2)<=(u-10^a/u)/2<=sqrt(2)*10^(a/2)
Multiply through by 2 and divide through by 10^(a/2).
2<=u/10^(a/2)-10^(a/2)/u<=2*sqrt(2)
Let v=u/10^(a/2). So v>1. Then:
2<=v-1/v<=2*sqrt(2).

We solve these two inequalities. First the left:
v-1/v>=2
v^2-2v-1>=0
v>=(1+sqrt(2)) or v<=(1-sqrt(2)).
v-1/v<=2*sqrt(2)
v>=(sqrt(2)+sqrt(3)) or v<=(sqrt(2)-sqrt(3)).
Since v>1, we drop the negative solutions and find:
1+sqrt(2) <= v <= sqrt(2)+sqrt(3).
or
1+sqrt(2) <= u/10^(a/2) <= sqrt(2)+sqrt(3).

We can do the same for y but we will find the same restriction on u.

Now we remember that u|10^a (u divides 10^a). Therefore u must be a power of
2 times a power of 5. Let u=5^b*2^c with b,c integers less than or equal to a.
Since we are going to divide it by 2, we must have c>=1.
Then we need to find a,b,c such that:
1+sqrt(2) <= 5^b*2^c/10^(a/2) <= sqrt(2)+sqrt(3)
These will give us u which will in turn determine x and y.
So take the log base 10 of all three sides. Since log is increasing, we do not
change the direction of inequality. Thus:
log(1+sqrt(2)) <= b*log(5)+c*log(2)-a/2 <= log(sqrt(2)+sqrt(3))
Multiply through by 2:
2*log(1+sqrt(2)) <= 2*b*log(5)+2*c*log(2)-a <= 2*log(sqrt(2)+sqrt(3))

If we approximate log(5) and log(2), this is sort of a Diophantine equation.
Since log(5) is very very close to 0.7 and log(2) is very very close to 0.3,
our approximations will be okay to find low solutions. If we want big solutions
then we need to use better convergents. We can calculate the boundary logs
as accurately as necessary. So:
0.77 <= 7/5*b+3/5*c-a <= 0.99
Multiply through by 5:
3.8 <= 7*b+3*c-5*a <= 4.9
So we must find a,b,c such that 7*b+3*c-5*a = 4, with a>b>=0 and a>=c>0.
There are many good ways to solve this but we will just pick a small solution.
b=3, c=1, a=4 (7*3+3-5*4=21+3-20=4)
Then u=5^3*2^1=250.
So y+x=250 and y-x=10^a/u=10^4/250=40.
Then y=145 and x=105.
y^2=21025 and x^2=11025.

This is, in fact, the smallest solution (it is easy to show that there is no
solution to the 7*b+3*c-5*a with a<4 and a>b>=0,a>=c>0).

>2.  What is the smallest square with leading digit 1 which remains a
>square when the leading 1 is replaced by a 2 and also remains a square
>when the leading digit is replaced by a 3?

We note from above that y=(5^b*2^c+10^a/(5^b*2^c)/2 or
2y=5^b*2^c+5^(a-b)*2^(a-c).

Should we now repeat the problem for a square with leading digit 2 that is
replaced by a 3, everything is the same except that y is now the smaller of the
pair. Thus:
2y=5^B*2^C-5^(a-B)*2^(a-C)
where B and C are different from b and c above but a is necessarily the same
(since we want the difference to be the same power of 10 for each transition).

Combining the two we get:
5^b*5^c+5^(a-b)*2^(a-b)=5^B*2^C-5^(a-B)*2^(a-C).
The proof that this has no solutions is too small to fit in the margin of this
posting.

>3.  What is the smallest square with leading digit 1 which remains a
>square when the leading 1 is replaced by a 2, and also remains a square
^S^Q^S^Q>when the leading digit is replaced by a 3, and also remains a square
>when the leading digit is replaced by a 4?
There is no solution since there is no solution to part 2.

Dan Cory


==> pickover/pickover.09.p <==
~Title: Cliff Puzzle 9: 3-Atoms and Growth
~From: [e-mail]

If you respond to this puzzle, if possible please send me your name,
address, affiliation, e-mail address, so I can properly credit you if
you provide unique information.  PLEASE ALSO directly mail me a copy of
your response in addition to any responding you do in the newsgroup.  I
will assume it is OK to describe your answer in any article or
publication I may write in the future, with attribution to you, unless
you state otherwise.  Thanks, Cliff Pickover

      * * *

    Start with 3 digits: 1, 2, and 3.
Each succeding row repeats the previous three rows, in order,
as you can see from the following diagram.

1
2
3
123
23123
312323123
12323123312323123
2312331232312312323123312323123

1. What is the sum of digits in the 100th row?

2. Get rid of all the twos.  Here I've replaced each of them with a "."

1
..
3
1.3
.31.3
31.3.31.3
1.3.31.331.3.31.3
.31.331.3.31.31.3.31.331.3.31.3

In the last row of this diagram, there are three different species:  31,
331 and 3.  How many different species are there in row 30?

3.  When the sequence first hits a three, it now undergoes an enzymatic
cleavage, and the digits on the right of the 3 are swapped with the
digits on the left.

1
2
3
123
23123 now becomes 12323
312312323 now becomes 123123233
Now answer the question posed in question 2.

==> pickover/pickover.09.s <==
-------------------------

~Subject: Re: Cliff Puzzle 9: 3-Atoms and Growth (PARTIAL SPOILER)
~Newsgroups: rec.puzzles
~References: <[e-mail]>

In article <[e-mail]>, Cliff Pickover writes:

> Start with 3 digits: 1, 2, and 3.
^S^Q^S> Each succeding row repeats the previous three rows, in order
> as you can see from the following diagram.
> 1
> 2
> 3
> 123
> 1. What is the sum of digits in the 100th row?

This one's easy.  You basically have a Tribonacci sequence with
the initial conditions S_1 = 1, S_2 = 2, S_3 = 3 and S_n = S_{n-1} +
S_{n-2} + S_{n-3} for n>3.  Thus, it's possible to find a closed
form of the type c_1*r_1^n + c_2*r_2^n + c_3*r_3^n.  Indeed, letting
T_i be the standard Tribonnaci sequence which has initial values
T_1 = 1, T_2 = 1, T_3 = 1 we can play a little game by noting the
T's go 1 1 1 3 5, and so by linearity S_i = ( T_i + T_{i+2} )/2, hence
S_100 = ( T_100 + T_102 )/2.
-------------------------


Dear Mr. Pickover,

I found your "123" problem interesting.  Here's the answers that I
came up with.  (Note: my personal info that you requested that I
include is at the end of the document.)

>      * * *

    >Start with 3 digits: 1, 2, and 3.
>Each succeding row repeats the previous three rows, in order,
>as you can see from the following diagram.

>1
>2
>3
>123
>23123
>312323123
>12323123312323123
>2312331232312312323123312323123

>1. What is the sum of digits in the 100th row?

Define an arithmetic series as follows:

(Note: Read a_1 as "a sub 1" and a_(n-1) as "a sub n-1".  I have
to do this because I can't use subscripts here.)

 a_1 = 1
 a_2 = 2
 a_3 = 3
 a_n = a_(n-3) + a_(n-2) + a_(n-1);  n>=4

The sum of each line is the sum of it's parts, so therefore, the
sum of each row is the sum of the previous three rows' sums.

a_30 = 45152016  (I wrote a simple basic program to calculate it.)

>2. Get rid of all the twos.  Here I've replaced each of them with a "."

>.31.3
>31.3.31.3
>1.3.31.331.3.31.3
>.31.331.3.31.31.3.31.331.3.31.3

>In the last row of this diagram, there are three different species:  31,
^Q^S^Q>331 and 3.  How many different species are there in row 30?

First, let me show that no "new" species will develop, other than those
seen in the sample few lines above:

        First, notice that there are four unique species above:
        "1","3","31","331".  Next, notice that the first species
        on a line goes in cycles of 3.  (Remember how we're building
        successive rows.  The first row repeated on a line is the
        row three back.  Hence the repeating pattern.)  Also notice
        that the ends of the rows do not change, this time because
        the last row represented on the current row is the row
        directly previous (and hence, it ends the same.)

        Because we are building successive rows via concatination,
        then only locations within new rows where "new" species may
        be found ("new" meaning not seen in any previous rows) is
        where the ends of two rows meet in the new row.  Since we
        know that the "end" of each row is limited to ".3" and
        the "beginnings" of each row cycle through "31", "1", ".",
        the only possible combinations we can make are "331", "31",
        and "3".  Since we alreadly have seen these, it is now
        obvious that we will create no more new species.
^S^Q
Next, let me show what species we WILL see:

        The species "3" is on the end of every line.  Therefore
        it will be in row 30.

        The species "31"  and the species "331" are both imbedded
        in a row previous to row 30.  Therefore they will be in
        row 30, because the "middle parts" of each row are
        duplicated down the list, not modified.

        The species "1" only shows up every third row.  It happens
        to occur on rows such that (Row #) mod 3 = 1.  Because
        30 mod 3=0, the species "1" will NOT occur in row 30.

        Hence, we have the three species "3","31","331" occuring
        in row 30.

>3.  When the sequence first hits a three, it now undergoes an enzymatic
>cleavage, and the digits on the right of the 3 are swapped with the
>digits on the left.

>1
>2
>3
>123
>23123 now becomes 12323
>312312323 now becomes 123123233
>Now answer the question posed in question 2.
I'm not taking the time to work this one out entirely.  It appears that
this algorithm forces 1's out in front all of the time, and keeps
appending 3's on the end of the row.  Hence, you'll see a proliferation
of species such as "3331","33331","333331", etc.  It also appears that
in row 30, you will have all the species from "3" , "31", "331","3331",
"33331", etc up to "33333333333333333333333331".  Now, I haven't
doublechecked my work here... I've been up all night, and am too
tired to double check my conjecture here.  But, I believe that I am
right, or at least on the right track.


I hope these answers help you.  I have two questions in return:
"Are you the 'pickover' responsible for many of the Fractint
fractal types?" and "Were my answers above even close?"  I apologize
if my answers seemed a little rough & non-formal at points.  I
hope you understand my explanation above.

Thanks for the mental workout.  I hope that this helps you, once again.

Hope to hear from you soon!

-- Joseph Zbiciak   [e-mail]

Here's that personal data to requested that I include:

I am Joseph Zbiciak, an Electrical/Computer Engineering Major at
Bradley University, Peoria, IL.   My current address is as follows:

Room 121, Heitz Hall
912 N Elmwood,
Peoria, IL  61606

My e-mail address is [e-mail].
Other info:  Year in school: Freshman,  DOB: 08/29/75
             Academic standing: good    Favorite toy: his computer
             Favorite hobby: spelunking through the internet looking
                                for tidbits like this question here.

If you need any more information, let me know.
Note:  I did not post this on the nn yet.  Feel free to for me, however.
        Thanks!


--
-------------------------

|> 3.When the sequence first hits a three, it now undergoes an enzymatic
|> cleavage, and the digits on the right of the 3 are swapped with the
|> digits on the left.
|>
|> 1
|> 2
|> 3
|> 123
|> 23123 now becomes 12323
|> 312312323 now becomes 123123233

>From how I understand the descriptive rule I get:

1
2
3
123 becomes 312
23123 becomes 12332
331223123 becomes 312231233

>From your example it seems that the trailing 3 is not regarded as a
'first' 3 (123 is not changed), nor is it regarded as a digit to be
swapped (as in the two other examples).
Is this how the rule should be interpreted?


And ... Keep up the good work, these are really good puzzles!!

--
[e-mail] (Norwegian Telecom Research)
 'When murders are committed by mathematics, they can be solved by
 mathematics. Most of them aren't, and this one wasn't'
 - Nick Charles (Dashiell Hammett's "The Thin Man")
-------------------------


Dear Dr. Pickover,

I found your "123" problem interesting.  Here's the answers that I
came up with.  (Note: my personal info that you requested that I
include is at the end of the document.)

>      * * *

    >Start with 3 digits: 1, 2, and 3.
>Each succeding row repeats the previous three rows, in order,
>as you can see from the following diagram.

>1
>2
>3
>123
>23123
>312323123
>12323123312323123
>2312331232312312323123312323123

>1. What is the sum of digits in the 100th row?

Define an arithmetic series as follows:

(Note: Read a_1 as "a sub 1" and a_(n-1) as "a sub n-1".  I have
^S^Q^Sto do this because I can't use subscripts here.)

 a_1 = 1
 a_2 = 2
 a_3 = 3
 a_n = a_(n-3) + a_(n-2) + a_(n-1);  n>=4

The sum of each line is the sum of it's parts, so therefore, the
sum of each row is the sum of the previous three rows' sums.

a_30 = 45152016  (I wrote a simple basic program to calculate it.)

>2. Get rid of all the twos.  Here I've replaced each of them with a "."

>.31.3
>31.3.31.3
>1.3.31.331.3.31.3
>.31.331.3.31.31.3.31.331.3.31.3

>In the last row of this diagram, there are three different species:  31,
>331 and 3.  How many different species are there in row 30?

First, let me show that no "new" species will develop, other than those
seen in the sample few lines above:

        First, notice that there are four unique species above:
        "1","3","31","331".  Next, notice that the first species
        on a line goes in cycles of 3.  (Remember how we're building
        successive rows.  The first row repeated on a line is the
        row three back.  Hence the repeating pattern.)  Also notice
        that the ends of the rows do not change, this time because
        the last row represented on the current row is the row
        directly previous (and hence, it ends the same.)

        Because we are building successive rows via concatination,
        then only locations within new rows where "new" species may
        be found ("new" meaning not seen in any previous rows) is
        where the ends of two rows meet in the new row.  Since we
        know that the "end" of each row is limited to ".3" and
        the "beginnings" of each row cycle through "31", "1", ".",
        the only possible combinations we can make are "331", "31",
        and "3".  Since we alreadly have seen these, it is now
        obvious that we will create no more new species.

Next, let me show what species we WILL see:

        The species "3" is on the end of every line.  Therefore
        it will be in row 30.
^Q^S^Q
        The species "31"  and the species "331" are both imbedded
        in a row previous to row 30.  Therefore they will be in
        row 30, because the "middle parts" of each row are
        duplicated down the list, not modified.

        The species "1" only shows up every third row.  It happens
        to occur on rows such that (Row #) mod 3 = 1.  Because
        30 mod 3=0, the species "1" will NOT occur in row 30.

        Hence, we have the three species "3","31","331" occuring
        in row 30.

>3.  When the sequence first hits a three, it now undergoes an enzymatic
>cleavage, and the digits on the right of the 3 are swapped with the
>digits on the left.

>1
>2
>3
>123
>23123 now becomes 12323
>312312323 now becomes 123123233
>Now answer the question posed in question 2.
I'm not taking the time to work this one out entirely.  It appears that
this algorithm forces 1's out in front all of the time, and keeps
appending 3's on the end of the row.  Hence, you'll see a proliferation
of species such as "3331","33331","333331", etc.  It also appears that

in row 30, you will have all the species from "3" , "31", "331","3331",
"33331", etc up to "33333333333333333333333331".  Now, I haven't
doublechecked my work here... I've been up all night, and am too
tired to double check my conjecture here.  But, I believe that I am
right, or at least on the right track.

Thanks for the mental workout.  I anxiously await more such puzzles!

Hope to hear from you soon!

-- Joseph Zbiciak   [e-mail]

Here's that personal data to requested that I include:

I am Joseph Zbiciak, an Electrical/Computer Engineering Major at
Bradley University, Peoria, IL.   My current address is as follows:

Room 121, Heitz Hall
B
912 N Elmwood,
Peoria, IL  61606

My e-mail address is [e-mail].
Other info:  Year in school: Freshman,  DOB: 08/29/75
             Academic standing: good    Favorite toy: his computer
             Favorite hobby: spelunking through the internet looking
                                for tidbits like this question here.


==> pickover/pickover.10.p <==
~Title: Cliff Puzzle 10: The Ark Series
~From: [e-mail]

If you respond to this puzzle, if possible please send me your name,
address, affiliation, e-mail address, so I can properly credit you if
you provide unique information.  PLEASE ALSO directly mail me a copy of
your response in addition to any responding you do in the newsgroup.  I
will assume it is OK to describe your answer in any article or
publication I may write in the future, with attribution to you, unless
you state otherwise.  Thanks, Cliff Pickover

      * * *

1.  Given a large ark containing 2 individuals of every animal species
in the world, what would be the approximate total weight of all the
organisms?  How would your answer differ if you included every plant,
bacterial, and fungal organism?

2.  Assume that all other organisms on earth were dead except for those
on the ark in question 1, and that the animals were released 1000 years
ago.  What would you expect to be surviving today?  (Assume that, where
applicable, a male and female were used for each species.)

3.  Assume that the year is 1992 and that it rained for 40 days, and the
^S^Q^S^Q^S^Q^S^Q^S^Q^Srain covered all the land on the earth.  Further assume that the flood
waters receded to pre-flood days within several months.

   What would be the geopolitical changes as a result of the
temporary flood?

   What would be the ecological changes as a result of the
temporary flood?

==> pickover/pickover.10.s <==
-------------------------

In article <[e-mail]> you write:
|> Title: Cliff Puzzle 10: The Ark Series
|> From: [e-mail]
|>
[ lotsa lines deleted ]
|>
|> 2.  Assume that all other organisms on earth were dead except for those
|> on the ark in question 1, and that the animals were released 1000 years
|> ago.  What would you expect to be surviving today?  (Assume that, where
                                                                     ^^^^^^^

|> applicable, a male and female were used for each species.)

Were you thinking of parthenogenesis or something ???
|>
|> 3.  Assume that the year is 1992 and that it rained for 40 days, and the
|> rain covered all the land on the earth.  Further assume that the flood
|> waters receded to pre-flood days within several months.
|>
|>    What would be the geopolitical changes as a result of the
|> temporary flood?

Dunno about this but it's a safe bet that the Netherlands _wouldn't_ get floode
d
We've been blocking the sea out for hundreds of years, so we've more experience
at it than anyone else.
|>
|>    What would be the ecological changes as a result of the
|> temporary flood?

Andy.

Just my opinions, nobody else's, especially not Oracle's
-------------------------

> 1.  Given a large ark containing 2 individuals of every animal species
> in the world, what would be the approximate total weight of all the
> organisms?  How would your answer differ if you included every plant,
> bacterial, and fungal organism?

1000 tons (guessed 10 million species with an average weight of 100 grams,
insects push this number down with their huge number of species).
No increase through bacteriae or fungi, but maybe with plants.
(You were unspecific:  All living species?)

> 2.  Assume that all other organisms on earth were dead except for those
> on the ark in question 1, and that the animals were released 1000 years
> ago.  What would you expect to be surviving today?  (Assume that, where
> applicable, a male and female were used for each species.)

None.  I think it's common knowledge with biologists that you need at least
~50 individuals of a species to keep genetic health --- aside from the
problem of both a male and female baby surviving.

> 3.  Assume that the year is 1992 and that it rained for 40 days, and the
> rain covered all the land on the earth.  Further assume that the flood
> waters receded to pre-flood days within several months.

"Covers the land."  How deep?  To cover *all* land (Himalaya) evenly, you
need a depth of 9000 m in most regions, so the question is, how fast will
it rise?  Do we just have time to put some tins in the boat?  Most people
don't have one.  Most airplanes cannot land but maybe some of them swim.
One has to calculate the distribution of swimming things in usual locations.
For if people have to swim 500-1000 m in cold water to a beam, most will
drown.

>    What would be the geopolitical changes as a result of the
> temporary flood?

With the survival of at most 1 percent of the population there will be a
completely new beginning.  Don't know if they would make the same mistakes,
though.  Technology will be thrown back, and science more than that.
Niven/Pournelle's "Lucifer's Hammer" is an accurate description.

>    What would be the ecological changes as a result of the
> temporary flood?

Lack of most animals, especially those dependent of plants (many of them
can't live without a day of food).  Most plants will grow again after some
time.

--ralf
  ************************************************************************
  After some tests, I decided to put 4 lines of sig here, because I really
  like the optical effect.  Now there's the problem what to write in it...
  ************************************************************************


==> pickover/pickover.11.p <==
~Title: Cliff Puzzle 11: The Leviathan Number
~From: [e-mail]

If you respond to this puzzle, if possible please send me your name,
address, affiliation, e-mail address, so I can properly credit you if
you provide unique information.  PLEASE ALSO directly mail me a copy of
your response in addition to any responding you do in the newsgroup.  I
will assume it is OK to describe your answer in any article or
publication I may write in the future, with attribution to you, unless
you state otherwise.  Thanks, Cliff Pickover

      * * *


   Many interesting observations have recently been published
concerning various number theory properties of the "number of the
beast", 666.  In this new
puzzle here I ask you to consider the monstrous
"leviathan number",
a number so large as to make the number of electrons,
protons, and neutrons in the universe (10**79) pale in comparison.  (It
also makes a googol (10**100) look kind of small).

The leviathan number is defined as (10**666)!, where the "!" indicates
factorial.

1.  What are the first 6 digits of the leviathan number?  Hint:  you
need not actually compute the leviathan to determine this.  If you can
determine the first 6 digits, please carefully spell out your method.

2. Could modern supercomputers compute the leviathan, or will this
beyond the realm of humankind for the next century?

3. Even if we cannot compute the leviathan, how many other
characteristics of this number can we write down.

==> pickover/pickover.11.s <==
-------------------------

~Subject: Re: Cliff Puzzle 11: The Leviathan Number (PARTIAL SPOILER)
~Newsgroups: rec.puzzles
~References: <[e-mail]>

In article <[e-mail]>, Cliff Pickover writes:

> The leviathan number is defined as (10**666)!, where the "!" indicates
> factorial.

> 1.  What are the first 6 digits of the leviathan number?

The simplest technique would be to use Stirling's formula to compute
the mantissa, i.e. frac( log(n) ) = frac( log(2*pi)/2 + log(n)/2
n*(log(n)-log(e)) ).  In our case n = 10^666, so this equals
frac( log(2*pi)/2 + 333 + 10^666*(666-log(e)) ) =
frac( log(2*pi)/2 + 10^666*(1-log(e)) ), so we'd basically need
to know something like 10 digits to the right of the decimal point
for log(2*pi)/2, and something like 700 digits for log(e) (which is
easily doable).  We then compute (1-log(e)), shift the digits 666
spaces to the left, and we're all set.

> 2. Could modern supercomputers compute the leviathan, or will this
> beyond the realm of humankind for the next century?

The number of digits is more than 10^668, and this compares
unfavorably to the number of particles in the universe.  Furthermore,
^Qeven if a googol digits could be output per second, you'd never
make it before the end of the universe.  So, I'd say it's beyond
the realm of humanity, period.

> 3. Even if we cannot compute the leviathan, how many other
> characteristics of this number can we write down.

As another puzzle, how many zeroes does it end with, and what are
the last two non-zero digits?
.qq
&EXIT
                    THIS FILE HAS BEEN RECEIVED FROM BIT^S^Q^SNET

          The file may be executable.  Before removing this header you
          must understand  what the code  will do. You must  also have
          the  appropriate intellectual  property agreements  in place
          before receiving the code into IBM.

          If you have any questions, contact your manager.


The contents of the file has been shifted right by one character.
   Filename=(none) Filetype=(none) RECFM=F LRECL=80 Records=21
The file received from the BITNET gateway begins below the next line.
------------------------------------------------------------------------
 Date:     Thu, 22 Oct 1992 07:12 EDT
 From:     <FRAMEM@UNION>
 Subject:  RE: googol!
 To:       CLIFF@YKTVMV
 Original_To:  Jnet%"CLIFF@YKTVMV"

 Hi, Cliff.

 The log10(e) comes from applying Stirling's approximation
 for the factorial: for large n, n! is approximately
 sqrt(2*pi*n)*((n/e)^n).  Substitute googol for n, take
 log10 of both sides, and recall the mantissa of the log10
 gives the digits of the original number.

 In these days of fast symbolic packages allowing exact
 computation of large factorials (though presumably not
 so large as a googol), people forget Stirling's formula.
 Until a few years ago, this was the only way to find
 factorials (albeit, only approximately) for large numbers.

 Mike

==> pickover/pickover.12.p <==
~Title: Cliff Puzzle 12: Slides in Hell
~From: [e-mail]

If you respond to this puzzle, if possible please send me your name,
address, affiliation, e-mail address, so I can properly credit you if
you provide unique information.  PLEASE ALSO directly mail me a copy of
your response in addition to any responding you do in the newsgroup.  I
will assume it is OK to describe your answer in any article or
publication I may write in the future, with attribution to you, unless
you state otherwise.  Thanks, Cliff Pickover

      * * *

Consider a metallic slide with 10 large holes in it equally spaced from
top to bottom.  If you attempt to slide down the slide you have a 50%
probability of sliding through each hole in the slide into an oleaginous
substance beneath the slide during each encounter with a hole.

1.  If you were a gambling person, which hole would you bet a person
would fall through?

2.  If you were a gambling person, how many attempts would it require
for a person to slide from the top of the slide to the bottom without
falling through a single hole.

3.  If all the people on earth lined up to go down the slide, and they
^Q^Sslid down a more horrifying slide with 100 holes at a rate of 1 person
per second, when would you expect the first person to arrive at the
bottom of the slide without falling through.
An hour? A day? A decade? ...
Received: from uoft02.utoledo.edu by watson.ibm.com (IBM VM SMTP V2R2) with TCP
;
~Title: Cliff Puzzle 12: Slides in Hell
>Consider a metallic slide with 10 large holes in it equally spaced from
>top to bottom.  If you attempt to slide down the slide you have a 50%
>probability of sliding through each hole in the slide into an
>oleaginous substance beneath the slide during each encounter with a
>hole.
>
>1.  If you were a gambling person, which hole would you bet a person
>would fall through?

None.  The best chance is the first hole but I got a 50-50 chance.  Why
bother?  (2nd hole is 1/4, 3rd 2**-3, ...)

>2.  If you were a gambling person, how many attempts would it require
>for a person to slide from the top of the slide to the bottom without
>falling through a single hole.

No gurantee.  Each slide is an independent event.  Now, if you are
talking mere probability, on the average, one in 1024 slides may make
it through all 10 holes.

>3.  If all the people on earth lined up to go down the slide, and they
>slid down a more horrifying slide with 100 holes at a rate of 1 person
>per second, when would you expect the first person to arrive at the
>bottom of the slide without falling through.  An hour? A day? A decade?

Again, can't tell.  It could be the first one, it could be none.  Probablity
can not foretell actual events.  But if you have infinite number of people
sliding down till eternity, on the average, you may see 1 person slide over
all holes every (2**100)/(365*24*69*6) years.  This number is many times
bigger than the world population for now.

==> pickover/pickover.12.s <==
-------------------------

In article <[e-mail]> you write:
: Consider a metallic slide with 10 large holes in it equally spaced from
^Q^S^Q: top to bottom.  If you attempt to slide down the slide you have a 50%
: probability of sliding through each hole in the slide into an oleaginous
: substance beneath the slide during each encounter with a hole.
:
: 1.  If you were a gambling person, which hole would you bet a person
: would fall through?
The chance of falling thru the first hole is 50%.  For the second hole, it
is (.5)(.5) = 25%, the thrid is (.5)^3 = .125.  The chance by the tenth
hole is about .0097 %.  Obviously, since I am limited to one hole, I would
place my money on hole #1 (best chance).

: 2.  If you were a gambling person, how many attempts would it require
: for a person to slide from the top of the slide to the bottom without
: falling through a single hole.
The sum of the prob for falling thru a hole is .5 + .5^2 + .5^3 +...+.5^10.
This is about 99.902% = .99902.  So about 98 times out of 100000, someone
will make it through without falling.  This is about 1 time out of 1020.
So give or take about 1020 tries....
:
: 3.  If all the people on earth lined up to go down the slide, and they
: slid down a more horrifying slide with 100 holes at a rate of 1 person
: per second, when would you expect the first person to arrive at the
: bottom of the slide without falling through.
: An hour? A day? A decade? ...
The prob for falling thru the last hole is .5^100 = 7.88x10^-31.  There must
be some chance less than this that one WILL make it thru the slide.  The MIN
number of tries that it must take is 1/.5^100 = 1.26x10^30.  At the given rate
this is about 9.647 x 10^23 years, much older than the universe if I remeber
correctly.
Also, the chance of making it must be GREATER than .5^101.  or with
all the math, the MAX amount of time is 1.929x10^24 years.  So give or
take about 1.5x10^24 years....



--
Michael Neylon   aka Masem the Great and Almighty Thermodynamics GOD!
      //         | Senior, Chemical Engineering, Univ. of Toledo
  \\ // Only the |   Summer Intern, NASA Lewis Research Center
\  \X/   AMIGA!  |         [e-mail]           /
 --------+ How do YOU spell 'potato'?  How 'bout 'lousy'? +----------
    "Me and Spike are big Malcolm 10 supporters." - J.S.,P.L.C.L
-------------------------

In rec.puzzles you write:

>Title: Cliff Puzzle 12: Slides in Hell
>From: [e-mail]
>
>If you respond to this puzzle, if possible please send me your name,
>address, affiliation, e-mail address, so I can properly credit you if
Jeff Rogers
Rensselaer Polytechnic institute
[e-mail]

>Consider a metallic slide with 10 large holes in it equally spaced from
>top to bottom.  If you attempt to slide down the slide you have a 50%
>probability of sliding through each hole in the slide into an oleaginous
>substance beneath the slide during each encounter with a hole.
>
>1.  If you were a gambling person, which hole would you bet a person
>would fall through?

The first one. There's only a 50% chance of them getting past it, and a
small chance of them falling into each succeeding hole.
hole #     percent chance of reaching and falling into
1            50
2            25
3            12.5
4            6.25
5            3.125
6            1.5625
7            0.78125
8            0.390625
^S^Q9            0.1953125
10           0.09765625

>
>2.  If you were a gambling person, how many attempts would it require
>for a person to slide from the top of the slide to the bottom without
>falling through a single hole.

The chances for reaching each succeeding hole are the same as reaching and
falling into the previous one. Therefore, the chances of passing all the
holes are the same as reaching and falling into the last hole  (see previous
answer for stats), which makes the probability .0009765625, so
statistically, 1024 slides would be required to guarantee reaching the
bottom. If I was a gambling person, I'd probably bet about half this,
because the actual events can happen in any order, and on average, I'd guess
that he'd get down in about 512 slides.

>
>3.  If all the people on earth lined up to go down the slide, and they
>slid down a more horrifying slide with 100 holes at a rate of 1 person
>per second, when would you expect the first person to arrive at the
>bottom of the slide without falling through.
>An hour? A day? A decade? ...

This is solved similarly; it is represented by powers of 2. To successfully
get past the last hole, it would require (statistically, at least) 2^100
or (by my trusty pocket calculator) 1.2676506 *10^30 slides.
More significant figures? dc! Which gives 1267650600228229401496703205376.
In similar logic as the last problem, I'd expect about half that, or
633825300114114700748351602688 slides. How much time would this be? Excluding
leap years, I calculate 20098468420665737593491 years. That's 20 sextillion
years, significantly more than the age of the universe, by about 11 orders
of magnitude. So I'd guess that no one will ever reach the bottom, they'll
all try and fail (assuming everyone only gets to go once), or die waiting in
line.

Diversion

--
"I can see 'em                          | "Want me to create a diversion?"
    I can see 'em                       | Diversion
        Someone wake me when it's over" | [e-mail]
^S^Q^S-------------------------

In article <[e-mail]> you write:
~Title: Cliff Puzzle 12: Slides in Hell
>Consider a metallic slide with 10 large holes in it equally spaced from
>top to bottom.  If you attempt to slide down the slide you have a 50%
>probability of sliding through each hole in the slide into an
>oleaginous substance beneath the slide during each encounter with a
>hole.
>
>1.  If you were a gambling person, which hole would you bet a person
>would fall through?

None.  The best chance is the first hole but I got a 50-50 chance.  Why
bother?  (2nd hole is 1/4, 3rd 2**-3, ...)

>2.  If you were a gambling person, how many attempts would it require
>for a person to slide from the top of the slide to the bottom without
>falling through a single hole.

No gurantee.  Each slide is an independent event.  Now, if you are
talking mere probability, on the average, one in 1024 slides may make
it through all 10 holes.

>3.  If all the people on earth lined up to go down the slide, and they
>slid down a more horrifying slide with 100 holes at a rate of 1 person
>per second, when would you expect the first person to arrive at the
>bottom of the slide without falling through.  An hour? A day? A decade?

Again, can't tell.  It could be the first one, it could be none.  Probablity
can not foretell actual events.  But if you have infinite number of people
sliding down till eternity, on the average, you may see 1 person slide over
all holes every (2**100)/(365*24*69*6) years.  This number is many times
bigger than the world population for now.
-------------------------

Some answers to your questions:

1. As the puzzle states there is a 50% chance of falling into each
hole, I would bet a person would fall into the first hole -- in a large
enough sample, 1/2 of the people will fall through the first hole, 1/4
through the second, 1/8 through the third, etc.

2. In a large sample, 1/(2^10) people would make it all the way down
the slide without falling through any of the holes (1/1024). This means
that 1023 out of 1024 people would fall through a hole. Using the
formula (1023/1024)^x=1/2, we can determine out of the first x people
to go down the slide, there is a 50% chance that one person will make
it down without falling through a hole.  The answer to this equation is
x=709.4 Thus I would bet that a person would make it all the way down
on one of the first 710 attempts.

3. As 2^100=1.2676*10^30 (roughly), and (including leaps years under
the Gregorian calendar) there are 31556952 seconds in the average year,
then statistically one person should make it down the slide every
4.017*10^22 YEARS. However, and this is a very rough estimate, I figure
the log of (1-1/(1.2676*10^30)) to be about -5.5*10^(-29). [I'm doing
the calculations on a scientific calculator which only has 10 places.]
Thus, using the formula xlog(1-1/2^100)=log(1/2), I get x=5.5*10^27.
Thus, there's about a 50% chance that after 5.5*10^27 seconds, someone
will have made it down the slide. To be on the safe side, I'd bet only
if I were given at least 6*10^27 seconds, a value which equals
1.901*10^20 YEARS.
^Q^S
I hope this answers the questions.

Ted Schuerzinger

email: [e-mail]
snailmail: HB 3819
Dartmouth College
Hanover, NH 03755
USA

In case you're wondering, I'm just a junior at Dartmouth who's
interested in puzzles like these. I'm not even a math major -- I'm a
double major in government and Russian.
-------------------------

In article <[e-mail]> you write:
>Title: Cliff Puzzle 12: Slides in Hell
>From: [e-mail]

>Consider a metallic slide with 10 large holes in it equally spaced from
>top to bottom.  If you attempt to slide down the slide you have a 50%
>probability of sliding through each hole in the slide into an oleaginous
>substance beneath the slide during each encounter with a hole.
>
>1.  If you were a gambling person, which hole would you bet a person
>would fall through?

There's a 50% chance of falling through the first hole, 25% the
second, 2^-n the n'th.  If the odds offered were the same, I'd go for
the first hole.

>2.  If you were a gambling person, how many attempts would it require
>for a person to slide from the top of the slide to the bottom without
>falling through a single hole.

You expect to make it 1 out of 1024 times; after 710 tries, the chance
of someone succeeding exceeds 1/2.  (Log base (1023/1024) of 1/2 is
709.4).

>3.  If all the people on earth lined up to go down the slide, and they
>slid down a more horrifying slide with 100 holes at a rate of 1 person
>per second, when would you expect the first person to arrive at the
>bottom of the slide without falling through.
>An hour? A day? A decade? ...

Never.  OK, 1/2^100 will make it.  There being under 2^33 people on
the planet, ...

After 4.2e22 years, the expected number of people who succeeded is 1;
after about 2.9e22 years, the chance of someone having succeeded is
about 1/2.

Like I said, never.

Seth            [e-mail]
^Q^S^Q-------------------------

In rec.puzzles you write:

>1.  If you were a gambling person, which hole would you bet a person
>would fall through?

If the pay-back odds were the same regardless of the hole, then obviously,
I'd bet on the first hole!  There's a 1:2 chance the person falls through
the first hole, a 1:4 combined chance of the person falling though the
second hole, etc...

>2.  If you were a gambling person, how many attempts would it require
>for a person to slide from the top of the slide to the bottom without
>falling through a single hole.

1024 is the median value for this case...  There's a 1:2**n chance of
a person falling through the nth hole, having missed all of the holes
before n.  Since the probability of falling through = the probability
passing over the hole safely (vs not ever getting there), the
probability that a person makes it to the end is also 1:1024.


>3.  If all the people on earth lined up to go down the slide, and they
>slid down a more horrifying slide with 100 holes at a rate of 1 person
>per second, when would you expect the first person to arrive at the
>bottom of the slide without falling through.
>An hour? A day? A decade? ...
There is a 1:2**(100-Log2(5 billion people)) chance that somebody makes
it through...  Given a finite # of people on the planet (approx 5 bil.)
I think we'll run out first...


--Joseph Zbiciak   [e-mail]


-------------------------

~Subject: Re: Cliff Puzzle 12: Slides in Hell (SPOILER)
~Newsgroups: rec.puzzles
~References: <[e-mail]>

In article <[e-mail]>, Cliff Pickover writes:

> Consider a metallic slide with 10 large holes in it equally spaced from
> top to bottom.  If you attempt to slide down the slide you have a 50%
> probability of sliding through each hole in the slide into an oleaginous
> substance beneath the slide during each encounter with a hole.

> 1.  If you were a gambling person, which hole would you bet a person
> would fall through?

The probability of falling into hole i is (1/2)^i, so your best bet
would be hole 1.

> 2.  If you were a gambling person, how many attempts would it require
> for a person to slide from the top of the slide to the bottom without
> falling through a single hole.

The probability of success is p = (1/2)^10, and as each trial is
independant the expected number of trials before success is 1/p or
2^10.

> 3.  If all the people on earth lined up to go down the slide, and they
^S^Q^S> slid down a more horrifying slide with 100 holes at a rate of 1 person
> per second, when would you expect the first person to arrive at the
> bottom of the slide without falling through.

In this case the number of expected trials is 2^100, which is much
larger than the total number of people.

> An hour? A day? A decade? ...

Try about 10^24 years.  As another problem, assuming a large enough
supply of sliders estimate when the slide will wear through from
friction.
-------------------------

In article <[e-mail]> you write:
>Title: Cliff Puzzle 12: Slides in Hell
>From: [e-mail]
>
>If you respond to this puzzle, if possible please send me your name,
>address, affiliation, e-mail address, so I can properly credit you if
>you provide unique information.  PLEASE ALSO directly mail me a copy of
>your response in addition to any responding you do in the newsgroup.  I
>will assume it is OK to describe your answer in any article or
>publication I may write in the future, with attribution to you, unless
>you state otherwise.  Thanks, Cliff Pickover
>
>      * * *
>
>Consider a metallic slide with 10 large holes in it equally spaced from
>top to bottom.  If you attempt to slide down the slide you have a 50%
>probability of sliding through each hole in the slide into an oleaginous
>substance beneath the slide during each encounter with a hole.
>
>1.  If you were a gambling person, which hole would you bet a person
>would fall through?

I'd bet that they fell through the first hole.  The probability of that
happening is 50%.  The probability of them falling through the second
hole is:
P(didn't fall through the first)*P(fell through the second) = 50%*50% = 25%

In general, P(falls through hole n)=
P(no fall through 1)*P(no fall through 2)*...*P(no fall through n-1)
 *P(fell through hole n).
For this problem, P(falls through hole n) is (50%)^n, where n is the hole #
from the top.

>2.  If you were a gambling person, how many attempts would it require
>for a person to slide from the top of the slide to the bottom without
>falling through a single hole.

(Hey, after the first failed attempt, they're screwed, no?)
P(success)=P(no fail)=P(no fall 1)P(no fall 2)...P(no fall 10)
 =50%^10
 =1/1024
They should make it at least one time in 1024.

>3.  If all the people on earth lined up to go down the slide, and they
>slid down a more horrifying slide with 100 holes at a rate of 1 person
>per second, when would you expect the first person to arrive at the
^Q^S^Q^S>bottom of the slide without falling through.
>An hour? A day? A decade? ...

Oh, one in about 4.02*10^22 years...  I wouldn't hold my breath.


-Richard
-------------------------

1. I would bet on the first hole, as there is a 0.5 probability of a person's
   falling into it, which is the highest such probability.

2. The probability of reaching the end of the slide on a particular try is
   1/2^10 = 1/1024.  In 709 tries, there is an approximately 0.5 probability of

3. Beats me - the even money bet is for a number of tries (approximately) equal
          ((2^100 - 1)/(2^100))
   calculate it.



--
_______________________________________________________________________
Dan Blum           Institute for the Learning Sciences   Room 327
[e-mail]   1890 Maple Ave., Evanston, IL 60201   708-467-2306

"Let it be granted that a controversy may be raised about any question,
 and at any distance from that question."
                                          Lewis Carroll
_______________________________________________________________________


==> pickover/pickover.13.p <==
~Title: Cliff Puzzle 13: Ladders to Heaven
~From: [e-mail]

If you respond to this puzzle, if possible please send me your name,
address, affiliation, e-mail address, so I can properly credit you if
you provide unique information.  PLEASE ALSO directly mail me a copy of
your response in addition to any responding you do in the newsgroup.  I
will assume it is OK to describe your answer in any article or
publication I may write in the future, with attribution to you, unless
you state otherwise.  Thanks, Cliff Pickover

      * * *

Consider the following scenario.  A standard ladder stretches from each
country on the earth upward a distance equal to the distance from the
earth to the moon.

Assume:
1. the ladder is made out of a strong metal such as
titanium, which will not break.
2. the ladder is inclined at a very steep angle, 70 degrees, for
each country.
3. there is a breathable atmosphere.
4. the people (or teams of people) are allowed to use standard
mountain climbing and camping gear, e.g. ropes, backpacks, etc. but not
sophisticated electrical mechanisms, engines, etc.
5. a reward is given to whomever reaches the top of the ladder
first: 1 million dollars to that person.  In addition the country's
national debt is wiped out.

Questions:
1.  Approximate how long it would take a person (or team of people) to
reach the top of the ladder.  Days?  Weeks?  Years?

2. Which country would be the first?
^Q^S
3. Is there any novel method you would suggest to achieve this goal?

4. Is this task impossible to carry out.

==> pickover/pickover.13.s <==
-------------------------

Interesting puzzle... Just one question though: Is there a moon,
i.e. is it possible to use the gravitational field of the moon to your
advantage by "falling upwards" once you have reached the point where
the moon's gravity is bigger than the erath's (and do we also assume that
the the climber(s) must survive the fall?? :-) or shall we assume that the
earth is alone in the universe?


Spyros Potamianos
[e-mail]
-------------------------

~Newsgroups: rec.puzzles
~Subject: Re: Cliff Puzzle 13: Ladders to Heaven
~References: <[e-mail]>
Organization: The Chrome Plated Megaphone of Destiny

>1.  Approximate how long it would take a person (or team of people) to
>reach the top of the ladder.  Days?  Weeks?  Years?

Note that after you're 22,300 miles from the earth's axis, you get to
"fall" the rest of the way, as long as you don't lose contact with
the ladder.

>2. Which country would be the first?

It has already been pointed out that countries on the equator have an
advantage.  I suppose you could consider that countries with a large
national debt have extra motivation.  :-)

>3. Is there any novel method you would suggest to achieve this goal?

I would suggest a bicycle-like vehicle clamped to the ladder.  By
pulling a light but strong rope on a pulley (perhaps obtained form
the same source as this fantastic ladder material), riders could be
changed fairly quickly, thanks to a crew of brawny pulley-pullers
with a variable-geared linkage to the rope.

For the rider to pull this ever-longer rope seems impossible, but I
think shorter segments could be lifted and linked.  Or the ground
crew could help the rider by pulling down rope from a hub of lesser
diameter than the wheels of the vehicle.

>4. Is this task impossible to carry out.

No.  I thought it might be impossible to halt at the far end of the
ladder and return, due to centrifugal acceleration, but that
acceleration turns out to be only about 5 cm/s^2.
__________________________________________________________
Matt Crawford       [e-mail]       Java Man


-------------------------

> How do we get food to the people?

I would have the riders change so often that they'd only need some
high-carbohydrate snacks and a couple quarts of fluid.  I think the
brawny ground crew could pull up the next rider, with his supplies
^Q^S^Qand another pulley and segment of rope, at an acceleration of about
0.5 g or better.  That would be under 90 minutes for each shift-
change up to the synchronous orbit level.

I haven't figured out yet how to link each new piece of rope that's
pulled up with a rider to the pulley that's at the high point reached
by the previous rider.  Linking is easy, but it would be nice to find
a way that lets the next pulled-up rider go from one segment to the
other without interruption.  Well, since the sky-buckets at
Disneyland do this trick at each end, I know it can be done.

I didn't know you'd written any books, but it was clear you're
working on one now.  Sure, send a list, but I have access to some
on-line catalogs, so maybe I can find them anyway.

                        Matt Crawford
-------------------------

> Consider the following scenario.  A standard ladder stretches from each
> country on the earth upward a distance equal to the distance from the
> earth to the moon.
>
> Assume:
> 1. the ladder is made out of a strong metal such as
> titanium, which will not break.
> 2. the ladder is inclined at a very steep angle, 70 degrees, for
> each country.
> 3. there is a breathable atmosphere.
> 4. the people (or teams of people) are allowed to use standard
> mountain climbing and camping gear, e.g. ropes, backpacks, etc. but not
> sophisticated electrical mechanisms, engines, etc.
> 5. a reward is given to whomever reaches the top of the ladder
> first: 1 million dollars to that person.  In addition the country's
> national debt is wiped out.

I would imagine that one would be able to fashion a hot air balloon given
condition 4.  Also, given condition 3, the hot air balloon would be able
to cover the entire distance.  One would then only need to attach a sliding
hookup between the ladder and the balloon and wait.

===M.Graf==[e-mail]==================================================


==> pickover/pickover.14.p <==
~Title: Cliff Puzzle 14: Geography Genuflection
~From: [e-mail]

If you respond to this puzzle, if possible please send me your name,
address, affiliation, e-mail address, so I can properly credit you if
you provide unique information.  PLEASE ALSO directly mail me a copy of
your response in addition to any responding you do in the newsgroup.  I
will assume it is OK to describe your answer in any article or
publication I may write in the future, with attribution to you, unless
you state otherwise.  Thanks, Cliff Pickover
^S^Q
      * * *

1.  How would the world be different today, geopolitically speaking, if
the ancient land masses had never drifted apart and, therefore,
today's world consisted of a single supercontintent?

2.  What would today's world be like if the land mass which formed the
Greek peninsula never existed?

3.  What would today's world be like if the land bridge which joined
Alaska to Asia never existed?

4.  Why do all the major peninsulas on earth point south?  See for
example:  Italy, Greece, Florida, and Baja, and the tips of Africa,
South America, India, Norway, Sweden, Greenland, and many other
landmasses.

==> pickover/pickover.14.s <==
-------------------------

In rec.puzzles you write:

>If you respond to this puzzle, if possible please send me your name,
>address, affiliation, e-mail address, so I can properly credit you if
>you provide unique information.
>
Mike Neergaard
University of Wisconsin
[e-mail]

I'm not a professional at this sort of thing, so I just summarized my
conclusions.  I'm sure they would be ripped to shreds by any competent
whatsit-type-individual-who-knows-all-about-this-kind-of-stuff.

>1.  How would the world be different today, geopolitically speaking, if
>the ancient land masses had never drifted apart and, therefore,
>today's world consisted of a single supercontintent?
We would all speak German.

>2.  What would today's world be like if the land mass which formed the
>Greek peninsula never existed?
>
We would know a low more about fluid dynamics.

>3.  What would today's world be like if the land bridge which joined
>Alaska to Asia never existed?
Christopher Columbus would be a national hero, instead of being vulnerable
to counter-claims of genocide.  America would have been settled several
decades later, due to a dearth of demonstrable natural resources.

>4.  Why do all the major peninsulas on earth point south?  See for
>example:  Italy, Greece, Florida, and Baja, and the tips of Africa,
>South America, India, Norway, Sweden, Greenland, and many other
>landmasses.
I just work here . . .
--
I really don't make any claim at all to know what I'm talking about.
Actually, I make no claim to know what YOU'RE talking about, either.
In fact, now I've forgotten what we were talking about . . .

-------------------------

In article <[e-mail]> you write:
>Title: Cliff Puzzle 14: Geography Genuflection
>From: [e-mail]
>
>If you respond to this puzzle, if possible please send me your name,
>address, affiliation, e-mail address, so I can properly credit you if
^S^Q^S^Q^S>you provide unique information.  PLEASE ALSO directly mail me a copy of
>your response in addition to any responding you do in the newsgroup.  I
>will assume it is OK to describe your answer in any article or
>publication I may write in the future, with attribution to you, unless
>you state otherwise.  Thanks, Cliff Pickover
>
>      * * *
>

Okay, administrative trivia first.  My name is Martin Eiger, you don't
need my address (home or business?), I don't want you citing my
affiliation if you quote me, and my e-mail address is
[e-mail].

>1.  How would the world be different today, geopolitically speaking, if
>the ancient land masses had never drifted apart and, therefore,
>today's world consisted of a single supercontintent?

My theory is that mankind would never have evolved.  The dominant
species would still be some sort of mammal, but not us.  This renders
a large number of geopolitical questions irrelevant.  For example,
elephant-like creatures are unlikely to care whether there is one or
two Germanys.


>2.  What would today's world be like if the land mass which formed the
>Greek peninsula never existed?

A tough one, since I'm not up on my Greek influences in the evolution
of civilization.  My guess is that civilization would have evolved
anyway, probably not too differently than it did.  It might not have
evolved as fast, i.e., we might now be where we were a thousand years
ago or so, but over the long haul, human history would follow a
similar course.


>3.  What would today's world be like if the land bridge which joined
>Alaska to Asia never existed?

Pretty much the same, I bet.  People would have colonized North
America anyway.  After all, they got to Hawaii, so somebody could
probably have gotten to North America.  And whether or not people
colonized North America from across the Pacific, people from Europe
would have paved the place over just the same.


>4.  Why do all the major peninsulas on earth point south?  See for
>example:  Italy, Greece, Florida, and Baja, and the tips of Africa,
>South America, India, Norway, Sweden, Greenland, and many other
>landmasses.

First of all, you have to define what's a major peninsula.  Secondly,
I don't like your list.  Norway and Sweden are on the same peninsula,
and Greenland is an island, not a peninsula.  And third, there are
plenty of perfectly fine peninsulas that don't point south:  Alaska,
Siberia, Michigan (two peninsulas for the price of one), Yucatan,
Arabia (points kind of southeast), and Iberia, for instance.  And
^Q^Sfourth, you missed a few good southern-pointing ones, such as Korea,
Crimea, the Sinai, and the one that kind of points from eastern
Siberia toward Japan that I'm sure has a name but I don't know it.  So
while there are lots of peninsulas pointing lots of directions, a
majority of them do seem to point south, and I have no idea why.

==> pickover/pickover.15.p <==
~Title: Cliff Puzzle 15: Cherries in Wine Glasses
~From: [e-mail]

If you respond to this puzzle, if possible please send me your name,
address, affiliation, e-mail address, so I can properly credit you if
you provide unique information.  PLEASE ALSO directly mail me a copy of
your response in addition to any responding you do in the newsgroup.  I
will assume it is OK to describe your answer in any article or
publication I may write in the future, with attribution to you, unless
you state otherwise.  Thanks, Cliff Pickover

      * * *

Consider a 9x9 grid of beautiful crystal wineglasses.  Throw 32 cherries
at the grid.  A glass is considered occupied if it contains at least one
cherry.  (With each throw a cherry goes into one of the glasses.)  How
many different patterns of occupied glasses can you make?  (A glass with
more than one cherry is considered the same as a glass with one cherry
in the pattern).

2.  Same as above except that you place 8 cherries in glasses (x,y) and
then determine the other positions by placing cherries at (x,-y),
(-x,y), (-x,-y) leading to 32 cherries in the grid.  Consider the array
of glasses centered at the origin.  How many different patterns of
occupied glasses can you make?  (A glass with more than one cherry is
considered the same as a glass with one cherry in the pattern).

3. Can your results be extrapolated to an NxN grid with M cherries
thrown at it for both problems?


==> pickover/pickover.15.s <==
In article <[e-mail]> you write:
^Q^S^Q: Consider a 9x9 grid of beautiful crystal wineglasses.  Throw 32 cherries
: at the grid.  A glass is considered occupied if it contains at least one
: cherry.  (With each throw a cherry goes into one of the glasses.)  How
: many different patterns of occupied glasses can you make?  (A glass with
: more than one cherry is considered the same as a glass with one cherry
: in the pattern).
Assuming that rotated patterns are allowed, then it is (simply)
sum( 81!/(81-n)! , n=1->32) . Since, if a total of n different classes are
filled, then the number of combinations is 81!/(81-n)!.  Since there can
be from 1 to 32 glasses filled, the total # is just the sum of these...

:
: 2.  Same as above except that you place 8 cherries in glasses (x,y) and
: then determine the other positions by placing cherries at (x,-y),
: (-x,y), (-x,-y) leading to 32 cherries in the grid.  Consider the array
: of glasses centered at the origin.  How many different patterns of
: occupied glasses can you make?  (A glass with more than one cherry is
: considered the same as a glass with one cherry in the pattern).
This limitation basically reduces the number of available spots, from 9x9
to 5x5.  Also, I only have to worry about 8 occupied spaces.  Soo...
#of comb. = sum( (25!/(25-n)!, n=1->8)
:
: 3. Can your results be extrapolated to an NxN grid with M cherries
: thrown at it for both problems?
With a odd N, and M = 4k (evenly divs by 4), then
for 1....
#of comb = sum( (N^2)!/(N^2-n)!  , n=1->M)
for 2....
#of comb = sum( (((N+1)/2)^2)!/(((N+1)/2)^2-n)! , n=1->M/4)

--
Michael Neylon   aka Masem the Great and Almighty Thermodynamics GOD!
      //         | Senior, Chemical Engineering, Univ. of Toledo
  \\ // Only the |   Summer Intern, NASA Lewis Research Center
\  \X/   AMIGA!  |         [e-mail]           /
 --------+ How do YOU spell 'potato'?  How 'bout 'lousy'? +----------
    "Me and Spike are big Malcolm 10 supporters." - J.S.,P.L.C.L

==> pickover/pickover.16.p <==
~Title: Cliff Puzzle 16: Undulating Squares
~From: [e-mail]

If you respond to this puzzle, if possible please send me your name,
address, affiliation, e-mail address, so I can properly credit you if
you provide unique information.  PLEASE ALSO directly mail me a copy of
your response in addition to any responding you do in the newsgroup.  I
will assume it is OK to describe your answer in any article or
publication I may write in the future, with attribution to you, unless
you state otherwise.  Thanks, Cliff Pickover

      * * *

A square number is of the form y=x**2.  For example, 25 is a square
number.

Undulating numbers are of the form:  ababababab... For example, the
following are undulating numbers:  1717171, 282828, etc.

1. Are there any undulating square numbers?

2. Are there any undulating cube numbers?


==> pickover/pickover.16.s <==
-------------------------

In article <[e-mail]> you write:
: 1. Are there any undulating square numbers?
11^2 = 121

: 2. Are there any undulating cube numbers?
7^3 = 343

(yes, I know they're short, but they qualify!)
^S^Q
--
Michael Neylon   aka Masem the Great and Almighty Thermodynamics GOD!
      //         | Senior, Chemical Engineering, Univ. of Toledo
  \\ // Only the |   Summer Intern, NASA Lewis Research Center
\  \X/   AMIGA!  |         [e-mail]           /
 --------+ How do YOU spell 'potato'?  How 'bout 'lousy'? +----------
    "Me and Spike are big Malcolm 10 supporters." - J.S.,P.L.C.L
-------------------------

In article <[e-mail]> you write:
>Hi, I was interested in non-trivial cases.  Those with greater
>than 3 digits.  Award goes to the person who finds the largest
>undulating square or cube number.  Thanks, Cliff

343 and 676 aren't trivial (unlike 121 and 484 it doesn't come from
obvious algebraic identities).  The chance that a "random"
number around x should be a perfect square is about 1/sqrt(x);
more generally, x^(-1+1/d) for a perfect d-th power.  Since
there are for each k only 90 k-digit undulants you expect
to find only finitely many of these that are perfect powers,
and none that are very large.  But provably listing all cases
is probably only barely, if at all, possible by present-day
methods for treating exponential Diophantine equations, unless
(as was shown in a rec.puzzles posting re your puzzles on
^S^Q^Sarith. prog. of squares with common difference 10^k) there is
some ad-hoc trick available.  At any rate the largest undulating
power is probably 69696=264^2, though 211^3=9393931 comes
remarkably close.

--Noam D. Elkies
-------------------------

In article <[e-mail]>, you write...
>1. Are there any undulating square numbers?
>
    Other than the obvious 11**2, 22**2, and 26**2, there is 264**2
which equals 69696.

>2. Are there any undulating cube numbers?
>
    Just 7**3 as far as I can tell, though I'm limited to IEEE computational
reals.

PauL M SchwartZ  (-Z-)  |  Follow men's eyes as they look to the skies
[e-mail]   |         the shifting shafts of shining
[e-mail]    |        weave the fabric of their dreams
[e-mail]    |                    - RUSH -


==> pickover/pickover.17.p <==
~Title: Cliff Puzzle 17: Weird Recursive Sequence
~From: [e-mail]

If you respond to this puzzle, if possible please send me your name,
address, affiliation, e-mail address, so I can properly credit you if
you provide unique information.  PLEASE ALSO directly mail me a copy of
your response in addition to any responding you do in the newsgroup.  I
will assume it is OK to describe your answer in any article or
publication I may write in the future, with attribution to you, unless
you state otherwise.  Thanks, Cliff Pickover

      * * *


Consider the simple yet weird recursive formula

a(n) = a(a(n-1)) + a(n-a(n-1))

The sequences starts with a(1) = 1, and a(2) = 1.  The "future" values
at higher values of n depend on past values in intricate recursive ways.
Can you determined the third member of the sequence?  At first, this may
seem a little complicated to evaluate, but you can being slowly, by
inserting values for n, as in the following:

a(3) = a(a(2)) + a(3-a(2))
a(3) = a(1) + a(3-1) =
a(3) = 1+1 = 2

Therefore, the 3rd value of the sequence a(3) is 2.

The sequence a(n) seems simple enough: 1, 1, 2, 2, 3, 4, 4, 4, 5, ...
Try computing a few additional numbers.  Can you find any interesting
patterns?  The prolific mathematician John H Conway presented this
recursive sequence at a recent talk entitled "Some Crazy Sequences."  He
noticed that the value a(n)/n approaches 1/2 as the sequence grows and n
becomes larger.  Can you find a value, N, above which the sequence the
value of a(n)/n is always within 0.05 of the value 1/2?  (In other
words,
.eq vbar a(n)/n -1/2 vbar lt 0.05.
The bars indicate the absolute value.)

   A difficult problem? you ask.
John Conway offered $10,000 to the person to find the s-m-a-l-l-e-s-t
such N. A month after Conway made the offer, Colin Mallows of AT&T
solved the $10,000 question:  N = 3,173,375,556.  Manfred Shroeder has
noted that the sequence is "replete with appealing self-similarities
that contain the clue to the problem's solution."  Can you find any
self-similarities?  As I write this, no-one on the planet has found a
value for the smallest N such that a(n)/n is always within 0.01 of the
value 1/2.
.eq (vbar a(n)/n -1/2 vbar lt 0.01. )



==> pickover/pickover.17.s <==
-------------------------

In article <[e-mail]> you write:
: Title: Cliff Puzzle 17: Weird Recursive Sequence
: Consider the simple yet weird recursive formula
: a(n) = a(a(n-1)) + a(n-a(n-1))

The first 32 terms, and the ratio a(n)/n for each is as follows...

n   a(n)     a(n)/n
1   1        1.0
2   1        1.0
3   2        .666
4   2        .5
5   3        .6
6   4        .666
7   4        .5714
8   4        .5
9   5        .5555
10  6        .6
11  7        .6363
12  7        .5833
13  8        .6153
^Q^S14  8        .5714
15  8        .5333
16  8        .5
17  9        .5294
18  10       .5555
19  11       .5789
20  12       .6
21  12       .5714
22  13       .5909
23  14       .6086
24  14       .5833
25  15       .6
26  15       .5769
27  15       .5555
28  16       .5714
29  16       .5517
30  16       .5333
31  16       .5161
32  16       .5
33  17 .... and so and....

off the top, we can see that on the 2^k (k a pos. int) terms, the
ratio goes to .5

between each of these, the ratio goes up and then drops back to .5
(ignoring the variances due to integer arithmatic)

the value of n at the maximum in each jump is halfway between the two
2^k points.  The value of a(n) at those points seems to be
2^(k-1) - f(k), where f(k) is some function that I cannot determine
without more computing power.... *sniff*

Therefore, we must find a value of x such that...
(2^(x-1)-f(x))/2^x - .5 <.05 (or whatever)
or
f(x)/2^x < .05

and then N would be .5*(2^x-2^(x-1))

if I could see the next terms up to 128, I might be able to calculate it...


--
Michael Neylon   aka Masem the Great and Almighty Thermodynamics GOD!
      //         | Senior, Chemical Engineering, Univ. of Toledo
  \\ // Only the |   Summer Intern, NASA Lewis Research Center
\  \X/   AMIGA!  |         [e-mail]           /
 --------+ How do YOU spell 'potato'?  How 'bout 'lousy'? +----------
    "Me and Spike are big Malcolm 10 supporters." - J.S.,P.L.C.L
-------------------------

In article <[e-mail]> you write:

>John Conway offered $10,000 to the person to find the s-m-a-l-l-e-s-t
>such N. A month after Conway made the offer, Colin Mallows of AT&T
>solved the $10,000 question:  N = 3,173,375,556.

As I pointed out in my posting, this is incorrect, and differs from
Mallows' correct answer published in his article. But a bit of
^Q^S^Qinvestigation shows that the above N is hardly a random guess, either.
Conway's sequence is best understood by analyzing it on "levels",
where the   k'th level is the set of integers between  2^k  and 2^(k+1).
It turns out that Mallows' correct answer, 6083008742, lies on level 32,
and the largest candidate answer on level 31 is  N=3173375556, the
number quoted above.

Where did you see the above value of N given as the answer to Conway's
question?

-tal    [e-mail]

p.s.  As I found out when I edited my posted response to your message,
      you either use lines longer than 80 characters in your postings,
      or else your editor appends extra linefeeds to each line.  Since
      both conditions could be problematic for a lot of people who read
      your messages on rec.puzzles, you might want to correct this
      condition.


==> pickover/pickover.18.p <==
~Title: Cliff Puzzle 18: Difficult Nested Roots
~From: [e-mail]

If you respond to this puzzle, if possible please send me your name,
address, affiliation, e-mail address, so I can properly credit you if
you provide unique information.  PLEASE ALSO directly mail me a copy of
your response in addition to any responding you do in the newsgroup.  I
will assume it is OK to describe your answer in any article or
publication I may write in the future, with attribution to you, unless
you state otherwise.  Thanks, Cliff Pickover

      * * *

Consider the following nested set of square roots.

.eq ? = sqrt <1 + G sqrt <1+(G+1) sqrt < 1 + ... >>>

Here, G indicates "Googol" or  10**100.
The "<" and ">" symbols indicate where the beginning and ends of the
the nested roots.

1. What is the value for in this infinite set of nested roots.
2. What is the next term under the root?

Hint:
In 1911, the famous mathematical prodigy Srinivasa Ramanujan posed the
following question (#298) in a new mathematical journal called the
:Journal of the Indian Mathematical Society.

.eq ? = sqrt <1 + 2 sqrt <1+3 sqrt <1 + ... >>>


==> pickover/pickover.18.s <==
-------------------------

In article <[e-mail]> you write:
: Title: Cliff Puzzle 18: Difficult Nested Roots
: From: [e-mail]
: Consider the following nested set of square roots.
:
:  ? = sqrt <1 + G sqrt <1+(G+1) sqrt < 1 + ... >>>
:
: Here, G indicates "Googol" or  10**100.
: The "<" and ">" symbols indicate where the beginning and ends of the
: the nested roots.
:
: 1. What is the value for in this infinite set of nested roots.
: 2. What is the next term under the root?
: Hint:
: In 1911, a twenty-three-year-old Indian clerk named Srinivasa Ramanujan
: posed the following question (#298) in a new mathematical journal called
: the Journal of the Indian Mathematical Society.
:
:  ? = sqrt <1 + 2 sqrt <1+3 sqrt <1 + ... >>>
:
Doing a n-depth thing-a-ding on this.....
n=1   v=1
2     1.732
3     2.236
4     2.5598
5     2.7551
6     2.867
....
20    2.99999376
....
so I expect that the sum is actually 3.  Or in the general case when the
2 (or the G from above) is replaced by m, then the evaluation of the series
is m+1.  This CAN be shown as follows....

m+1 = sqrt(1+m sqrt(1+(m+1)*sqrt(....))
m^2 + 2m +1 = 1 + m *sqrt(1 + (m+1)*sqrt(...))
m^2 + 2m = m*sqrt(1+(m+1)*sqrt(...))
^S^Qm+2 = sqrt(1+(m+1)*sqrt(1+(m+2)*sqrt(...))

Thus if m+1 is then sum when the series is based off m, then m+2 is then
sum when the series is based off m+1.  Since this works for m=2 (as shown
above), then it must work for all whole numbers (mathematical induction is
such a wonderful thing...)

Therefore, the sum with m=G is G+1.

The next term, as show above, is (1+(m+2)*sqrt(1+....))


--
Michael Neylon   aka Masem the Great and Almighty Thermodynamics GOD!
      //         | Senior, Chemical Engineering, Univ. of Toledo
  \\ // Only the |   Summer Intern, NASA Lewis Research Center
\  \X/   AMIGA!  |         [e-mail]           /
 --------+ How do YOU spell 'potato'?  How 'bout 'lousy'? +----------
    "Me and Spike are big Malcolm 10 supporters." - J.S.,P.L.C.L