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### More Monkeys and Nuts

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Date: 06/07/99 at 20:29:17
From: Michael Holt
Subject: Further monkey puzzling - point raised

Dear Dr(s) Math,

I have looked through your archive and found a couple of variants on
the monkeys sharing nuts theme. None is specifically the same as the
problem I have (I don't think) although it is very similar to one
that is answered at some length with formulae and substitution, etc.
The person who set me this puzzle suggested that it was possible to
deduce the answer without going to this trouble, and gave me a clue
which I will add at the end of the puzzle.

The puzzle:

Five monkeys collect nuts, which they leave in a pile. During the
night one wakes up and divides the pile into 5. There is one left
over. He eats the remainder nut and his 1/5 share, and mixes the nuts
together. The second monkey comes down, divides the pile into 5 with
one left over, and so it goes on; each left pile divides into 5 with
one left over and each monkey eats its share plus one. The question
is, what is the least number of nuts that they had amassed in order
for this to work?

The clue:

Apparently, by asking what would have happened if there were 4 more
nuts collected, the answer can be found without lengthy calculations -
in fact I am told of somebody who solved it in his head over lunch
given this advice. I have played around with the idea at length but
have grown frustrated, as I have not yet solved the problem. What do
you think?

Thanks for your help; I look forward to hearing from you.

Yours sincerely,
Michael Holt
```

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Date: 06/10/99 at 12:42:20
From: Doctor Peterson
Subject: Re: Further monkey puzzling - point raised

Hi, Michael. I've been playing with this off and on for a few days,
and getting confused because of the variations among versions of the
puzzles! My biggest frustration has been trying to make my answer
agree with one of the others.

Yes, this clue does help, at least with your version of the puzzle.

Let's suppose that the initial number of nuts is X_0. I divide the
pile by five and have one left over, so each pile is (X_0-1)/5, and
the number I leave is

X_1 = (X_0 - 1) * 4/5

The same process is repeated five times until I have a fairly
complicated expression for X_5, the number left at the end.

Suppose we set 4 nuts next to the pile and don't touch them through
the whole process (so they AREN'T part of the dividing by five -
although that's not quite how you described the hint). The total
number of nuts is

Y_0 = X_0 + 4

Y_1 = X_1 + 4
= (X_0 - 1) * 4/5 + 4
= (Y_0 - 4 - 1) * 4/5 + 4
= (Y_0 - 5) * 4/5 + 4
= Y_0 * 4/5 - 4 + 4
= Y_0 * 4/5

So at each step, the total number of nuts is just multiplied by 4/5,
and we can easily see that

Y_5 = Y_0 * (4/5)^5

For this to be a whole number, Y_0 must be a multiple of 5^5, and the
smallest possible value would be exactly that, or 3125. Then X_0 is
four less than this, or 3121.

Now let's try it out:

X_0 = 3121
X_1 = (3121 - 1) * 4/5 = 2496
X_2 = (2496 - 1) * 4/5 = 1996
X_3 = (1996 - 1) * 4/5 = 1596
X_4 = (1596 - 1) * 4/5 = 1276
X_5 = (1276 - 1) * 4/5 = 1020

This final number, of course, is 4^5 - 4.

This problem is NOT the same as

http://mathforum.org/dr.math/problems/koestler12.18.97.html

where the recursion equation is

X_1 = X_0 * 4/5 - 1

and there is an extra round at the end in which the pile is divided
into five openly. I don't know whether there's a similar trick here.

Your problem IS, however, the same as

http://mathforum.org/dr.math/problems/stillman10.12.98.html

except that there we have the extra round, so the answer using our
method should be 5^6 - 4 = 15621, and the number each one gets at the
end should be 4^6 - 4 = 4092.

The equation found there is

1024*a - 15625*g = 11529

Plugging in our values for a = X_0 and g = X_6/4 (the number each gets
at the end), we see

1024*15621 - 15625*1023 = 15995904 - 15984375 = 11529

and we have it.

The trick amounts to a change of variables that simplifies the
equation. To find such a change, you could try Y = X + a, and see what
value of a will help:

X_1 = (X_0 - 1) * 4/5
Y_1 - a = (Y_0 - a - 1) * 4/5
Y_1 = Y_0 * 4/5 + a - (a + 1)*4/5

To make the equation simpler, we want

a - (a + 1)*4/5 = 0
5a = 4(a + 1)
a = 4

Or, you can just have a flash of insight and try adding four nuts for
no reason at all.

- Doctor Peterson, The Math Forum
http://mathforum.org/dr.math/
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High School Puzzles
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