Monkeys, Coconuts, and Seven Piles

Date: 11/29/2001 at 12:13:07
From: mike
Subject: lots of equations, lots of unknowns...

I've tried to work this problem lots of ways, but it has so many 
equations that I'm never sure if I'm getting it right.

Seven monkeys spend the day gathering coconuts. As night falls, they 
gather all the coconuts into one pile and agree to divide them up 
evenly in the morning. 

At 12:00 midnight one monkey gets up and divides the pile of coconuts 
into seven piles with one left over, which he throws into the ocean.  
He takes one of the piles and hides it in a secret hiding place and 
combines the remaining six for the others to find in the morning. At 
1:00 A.M., another monkey wakes up and divides the pile into seven, 
throwing the extra one into the ocean, hiding one pile in his secret 
hiding place, then grouping the rest. 

All seven monkeys do this. In the morning, they get up to divide the 
pile and it looks noticeably smaller! They divide the pile into seven, 
throw the left over one into the ocean, then go home. What is the 
least number of coconuts they could possibly have started with, and 
how many coconuts did each monkey get?

7a+1 = 6b
7b+1 = 6c
7c+1 = 6d
7d+1 = 6e
7e+1 = 6f
7g+1 = 6g
7h+1 = T

OR

a = starting amount of coconuts
b = (6/7)(a-1)
c = (6/7)(b-1)
d = (6/7)(c-1)
e = (6/7)(d-1)
f = (6/7)(e-1)
g = (6/7)(f-1)
h = (6/7)(g-1)
i = (6/7)(h-1)
i = 7j+1

I get messed up making the equation, then also finding an integer for 
how many coconuts each monkey got in the end.

Date: 11/29/2001 at 16:34:18
From: Doctor Rob
Subject: Re: lots of equations, lots of unknowns...

Thanks for writing to Ask Dr. Math, Mike.

This is very similar to the following problems from the Doctor Math
archives:

   Monkeys Dividing the Coconut Pile
   http://www.mathforum.org/dr.math/problems/thomas7.28.98.html   

   Coconut and Monkey Puzzle
   http://www.mathforum.org/dr.math/problems/koestler12.18.97.html   

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

   More Monkeys and Nuts
   http://www.mathforum.org/dr.math/problems/holt.6.07.99.html   

The equations you have to solve are

   a = starting amt of coconuts
   b = (6/7)*(a-1) = amt after 1 monkey,
   c = (6/7)*(b-1) = amt after 2 monkeys,
   d = (6/7)*(c-1) = amt after 3 monkeys,
   e = (6/7)*(d-1) = amt after 4 monkeys,
   f = (6/7)*(e-1) = amt after 5 monkeys,
   g = (6/7)*(f-1) = amt after 6 monkeys,
   h = (6/7)*(g-1) = amt after 7 monkeys,
   i = (1/7)*(h-1) = amt each got at the end.
   
Clearing fractions, these become

   6*a - 7*b = 6,
   6*b - 7*c = 6,
   6*c - 7*d = 6,
   6*d - 7*e = 6,
   6*e - 7*f = 6,
   6*f - 7*g = 6,
   6*g - 7*h = 6,
     h - 7*i = 1.

Eliminating in turn b, c, d, e, f, g, and h, these equations give,
in turn,

   6^1*a - 7^1*b = 6*(7^1 - 6^1),
   6^2*a - 7^2*c = 6*(7^2 - 6^2),
   6^3*a - 7^3*d = 6*(7^3 - 6^3),
   6^4*a - 7^4*e = 6*(7^4 - 6^4),
   6^5*a - 7^5*f = 6*(7^5 - 6^5),
   6^6*a - 7^6*g = 6*(7^6 - 6^6),
   6^7*a - 7^7*h = 6*(7^7 - 6^7),
   6^7*a - 7^8*i = 7^8 - 6^8.

This last equation has a particular solution a = -6 and i = -1. (Check 
to make sure you see why this works.) The general solution is given by

   a = -6 + 7^8*t,
   i = -1 + 6^7*t,

where t can be any integer. Putting t = 1 gives you the smallest 
possible positive answer for a and i. I leave it to you to find the 
number of coconuts each monkey got.

With over 5 million coconuts at the start, there was a whole lot
of counting going on during the night!

Notice the really nice "virtual solution" starting with a = -6 
coconuts:  a = b = c = d = e = f = g = h = -6, i = -1, each
monkey ends up with -2 coconuts, and 8 end up in the ocean!

Feel free to write again if I can help further.

- Doctor Rob, The Math Forum
  http://mathforum.org/dr.math/