by Bill Graham

Begin by restating this problem in the language of mathematics. The initial number of coconuts is N. After the first sailor does his or her bit, the number left is N' = 2(N-1)/3. This process is repeated three times altogether leaving an even multiple of three coconuts that is less than 10.

Only 9, 6, 3, and 0 can be left. The most direct method for solution will try each number out to see if it works.

Note that doing the sailor thing backward means that you add (back) half again as many coconuts as you have and then add one more. You cannot add one-half of 3 or 9. Thus, only 0 and 6 remain as finishing points. If you choose 0, then you'll have 1 after a single step and will be unable to make another step because you cannot divide 1 in half and still have whole coconuts.

Take 6, add 1/2 again as much to get 9, and add 1 to get 10. Put this into a table.

6	*1.5 = 9	+1 = 10
10	*1.5 = 15	+1 = 16
16	*1.5 = 24	+1 = 25

Looks like the sailors started out with 25 coconuts in the original pile.

Think some more
Removing the "fewer than 10" constraint should create some new possibilities. Let the final amount of coconuts be 3n, where n is an integer. Redo the table.

3n	*3/2 = 9n/2	+1 = 9n/2+1
9n/2+1	*3/2 = 27n/4 + 3/2	+1 = 27n/4 + 5/2
27n/4 + 5/2	*3/2 = 81n/8 + 15/4	+1 = 81n/8 + 19/4

The initial number of coconuts is 81n/8 + 19/4 or (10-1/8)n + 4-3/4 and must be an integer. For this reason, n/8 must be a number whose fractional part is 1/4. The smallest number, n, that does this is 2, which gives a final value of 3n = 6 and an initial number of 25 in excellent agreement with the solution above. The table below enumerates some additional solutions, each of which requires that you add 8 to the previous value of n.

n	final amount	initial amount
2	6	25
10	30	106
18	54	187
26	78	268
34	102	349

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Contents copyright 2004 by Bill Graham and ParaComp, Inc. All rights reserved.
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