Monkeys Dividing the Coconut Pile

Date: 07/28/98 at 22:44:37
From: Thomas
Subject: Monkey Problem

This is a problem I could not solve. Please help me out. 

Three monkeys spend a day gathering coconuts together. When they have 
finished, they are very tired and fall asleep. 

The following morning the first monkey wakes up. Not wishing to disturb 
his friends, he decides to divide the coconuts into 3 equal piles, but 
there is one left over, so he throws this odd one away, helps himself 
to his share, and goes home. 

A few minutes later the second monkey awakes. Not realizing that the 
first has already gone, he too divides the coconuts into 3 equal heaps, 
finds one left over, throws the odd one away, helps himself to his fair 
share, and goes home. 

Then the third monkey does exactly the same. 

What is the smallest possible number of coconuts left?" 

Many thanks,
Thomas

Date: 07/29/98 at 11:10:16
From: Doctor Anthony
Subject: Re: Monkey Problem

If N = original number of coconuts, then the first monkey takes 
(N-1)/3 as his share, leaving 2(N-1)/3 behind. Then the second monkey 
takes:

   2(N-1)/3 - 1    2(N-1)    1
   ------------  = ------ - --- 
        3             9      3

leaving behind:

   4(N-1)    2  
   ------ - ---   
     9       3

The third monkey takes:

   4(N-1)/9 - 2/3 - 1      4(N-1)    5
   -------------------  =  ------ - --- 
           3                 27      9

leaving:

   8(N-1)    10       8(N-1) - 30
   ------ -  ---  =  -------------       (Equation 1)
     27       9            27

Here, we assume that the third monkey does not just grab everything 
left after the other two have gone.

We require Equation 1 to be an integer, say t, and so:

   8(N-1) - 30 = 27t

   8N = 38 + 27t      where both N and t are integers.

   8N - 27t = 38   

Then divide through by 8, the smaller coefficient:

   N - 3t - 3t/8 = 4 + 6/8

This equation must be satisfied in integers. This means that the 
fractions in the above equation must reduce to an integer. Thus:

    3t   6   3t+6
   --- + - = ---- = integer 
    8    8    8

To find the smallest value, we can start by testing whether t=1 results 
in an integer and then increasing t by 1 until we get to the first 
integer. For example, if t = 1, then (3(1) + 6)/8 = 9/8, not an 
integer. If t = 2, then (3(2) + 6)/8 = 12/8 = 3/2, not an integer. 
We can continue this process until we get to the first integer. This 
may take a while, or it may not. 

Alternatively, we could use the following method. First, multiply the 
top line by 3. This is because we want the coefficient of t in the 
numerator to be 1. So:

   9t + 18
   ------- = integer
      8 

Thus:

   t + (t + 18)/8 = integer  

So we want:

   (t + 18)/8 = integer = p

   t+18 = 8p

   t = 8p - 18

Since t must be positive, we take p = 3 and get t = 24-18 = 6.

So the minimum number of coconuts left by the third monkey is 6.
              
- Doctor Anthony, The Math Forum
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