Coconuts, Forwards and Backwards

Date: 02/02/2010 at 13:09:09
From: sue
Subject: x/3 +1 +(x/3 +1)/3 + 2/3 +(x/3) + 3 =22

Three sailors were marooned on an island.  They worked all day 
collecting coconuts, which left them too tired to settle up that 
night; so they decided to divide the coconuts in the morning.

In the night, one sailor woke up and decided to separate his share.  
He was able to make three piles, with one coconut left over, which he 
threw away.  He put his share away and left the remainder in a single 
pile.

Later that night, the second sailor awoke and did the same thing, 
with one left over that he threw away.  Later still, the third sailor 
awoke and did the same thing with the remaining coconuts.

In the morning, the three sailors noticed the pile was smaller, but 
said nothing.  They divided what was left of the original pile 
equally.  Each sailor received seven coconuts, and one was left over, 
which they threw away.

How many coconuts were in the original pile?

I can start at the end and work back, but after the 22 from the end, 
I get confused as to how to get the other numbers.

Date: 02/02/2010 at 23:01:16
From: Doctor Greenie
Subject: Re: x/3 +1 +(x/3 +1)/3 + 2/3 +(x/3) + 3 =22

Hi, Sue --

This is a classic problem.  There are several pages in the Dr. Math 
archives where similar problems are discussed and solved; you can 
find links to those pages by searching the archives using the keyword 
"coconut."

However, most of those explanations get into some REALLY ugly 
algebra.  So let's see if I can explain how to get to the answer to 
your problem with less work.

You can't avoid a mess if you decide to work the problem "forward" -- 
that is, starting with x as the original number of coconuts, and 
finding algebraic expressions for the number remaining at each point 
in the story.  But let's look at that process first, and then see how 
much easier it is to work the problem backwards.

Whether we work the problem forward or backward, we want to work with 
the number of coconuts remaining at each point, rather than thinking 
about the number each sailor takes.

The same thing happens three times: the number of coconuts remaining 
is 1 more than a multiple of 3, so 1 coconut is thrown away; then a 
sailor takes one-third of the remaining coconuts and leaves the other 
two-thirds of the remaining coconuts in the pile.  Here is how that 
looks algebraically:

             x = original number of coconuts
         x - 1 = the result of throwing 1 away, which also 
                    makes the remaining number divisible by 3
  (1/3)(x - 1) = number the sailor takes
  (2/3)(x - 1) = number remaining after the sailor takes his share

So at each stage, the number of leftover coconuts is

(1) reduced by 1; and then
(2) multiplied by 2/3

If we take that rule -- "subtract 1, then multiply by 2/3" -- and 
apply it three times in succession, starting with the original number 
"x" of coconuts, we get the following:

original number:  x

1st sailor:

       subtract 1:  x - 1
  multiply by 2/3:  (2/3)(x - 1)
                  = (2/3)x - (2/3)

2nd sailor:

       subtract 1:  (2/3)x - (2/3) - 1
                  = (2/3)x - 5/3
  multiply by 2/3:  (2/3)[(2/3)x - 5/3]
                  = (4/9)x - (10/9)

3rd sailor:

       subtract 1:  (4/9)x - (10/9) - 1
                  = (4/9)x - 19/9
  multiply by 2/3:  (2/3)[(4/9)x - 19/9]
                  = (8/27)x - (38/27)

The problem tells us the number of coconuts remaining after the third 
sailor took his share: 22.  So we set that equal to our last 
expression, above:

  (8/27)x - (38/27) = 22
            8x - 38 = 22*27
            8x - 38 = 594
                 8x = 632
                  x = 79

The number of coconuts originally in the pile is 79.

We can check this answer by working through the problem one step at a 
time:

  start: 79
  1st sailor throws away 1; remaining are 79 - 1 = 78
  1st sailor takes 78/3 = 26; remaining are 78 - 26 = 52
  2nd sailor throws away 1; remaining are 52 - 1 = 51
  2nd sailor takes 51/3 = 17; remaining are 51 - 17 = 34
  3rd sailor throws away 1; remaining are 34 - 1 = 33
  3rd sailor take 33/3 = 11; remaining are 33 - 11 = 22

In the preceding analysis, we used the rule "subtract 1; multiply by 
2/3" to get from the number of coconuts at one point in the story to 
the number of coconuts at the next similar point in the story.  This 
rule acts like a function:

  f(x) = (2/3)(x - 1)

Applying this function three times in succession, starting with our 
answer of 79 coconuts in the original pile, leads us to 22 as the 
number of coconuts in the pile after all three sailors have taken 
theirs:

  f(79) = (2/3)(79 - 1) = (2/3)(78) = 52
  f(52) = (2/3)(52 - 1) = (2/3)(51) = 34
  f(34) = (2/3)(34 - 1) = (2/3)(33) = 22

Now, we needed some relatively ugly arithmetic to use the preceding 
function.  We also needed to know that after three steps, only 22 
coconuts were left; and needed to declare "x" as the original number 
of coconuts, and then use that variable in an equation that we could 
solve to find that the original number of coconuts was 79.

But we can use the INVERSE of our function to work BACKWARDS, using 
the KNOWN final number of coconuts, 22, to calculate the original 
number of coconuts without having to do ANY algebra.

The function we used to work "forwards" through the problem was

  (1) subtract 1;
  (2) multiply by 2/3

The inverse function performs the opposite operations in the opposite 
order:

  (1) divide by 2/3  (that is; multiply by 3/2);
  (2) add 1

We can start with the known final number of 22 coconuts and apply 
this inverse function three times to find the original number of 
coconuts in the pile:

  end:  22
  before the 3rd sailor:  (3/2)22 + 1 = 33 + 1 = 34
  before the 2nd sailor:  (3/2)34 + 1 = 51 + 1 = 52
  before the 1st sailor:  (3/2)52 + 1 = 78 + 1 = 79

Just a BIT easier than all the ugly algebra we needed to solve the 
problem working "forwards"....!!

I hope this helps.  Please write back if you have any further 
questions about any of this.

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