Chapter 1: The Monkey and the Coconuts

The Monkey and the Coconuts

In this chapter, Ben Ames Williams proposed a 5 sailor and 1 monkey problem involving coconuts. Today, the problem of coconuts is probably the most worked on and the least often solved of Diophantine brainteasers (equations calling for solutions in rational number), mentioned by Gardener in the chapter. In simple terms, there are 5 sailors and 1 monkey who are stranded and collected a pile of coconuts. Each man woke up during the night and separated the pile into fifths with having one left over of which they gave to the monkey. In the morning they separated what was left and came up with equal piles. The question is, “How many coconuts were there in the beginning?” This problem can seem rather simple, but, it is quite the opposite. This problem has an infinite amount of answers in whole numbers, but in the chapter, our job is to find the smallest positive number.

This problem can be expressed with these indeterminate equations, which represent the successive divisions of the coconuts into fifths. N is the original coconut number and F is the number each sailor received on the final division and the +1’s are the coconuts tossed to the monkey.

N=5A +1,

4A=5B+1,

4B=5C+1,

4C=5D+1,

4D=5E+1,

4E=5F+1.

In reduced form, using algebraic methods this can be simplified to:

1,024N=15,625F + 11,529.

Gardner states the equation is too difficult to prove from trial and error so he uses an alternative method of instead solving this with negative coconuts! Since N is divided six times into five piles, 5⁶ (or 15,625) can be added to any answer to give us the next answer. We could also do this method for finding infinite numbers in the negatives, these numbers will satisfy the equation but not the actual problem because the answer has to be a positive integer.

The negative number he uses to solve the equation is -4. The sailor approaches the   -4 coconut pile and tosses a positive coconut to the monkey, which leaves -5 coconuts now; he divides the pile into fifths, takes his pile and hides it. This method leaves us with exactly -4 coconuts again! When all the sailors finish they have -2 coconuts (because they split the pile at the end) and the monkey has +6 coconuts. Now to obtain the solution, we just subtract 4 from 15,625 to get 15,621 as our solution. This approach to the problem gives us the ability to solve different coconut problems with different amounts of sailors equal to k(nⁿ⁺¹) – m(n-1) where n is men, m is number of coconuts and k is an arbitrary integer called the parameter.

In Dividing Coconuts: A Linear Diophantine Problem from S. Singh and D. Bhattacharya goes on to explain the problem in greater depth. They give two solutions to the problem using alternative solving methods and by writing out the actually process to obtain the solution. The math was shown for the solution of the   -4 coconut solution Gardner explained which Gardner chose to leave out in the chapter. What interested me was the alternative method they used, even though I hardly understood it.

In Monkeys and Coconuts by N. Anning, she too goes into further depth of solving the coconut problem. N Anning, many gave examples of the coconut problem and solving it with different amounts of sailors and monkeys. She also goes on to talk about the original monkey and coconut problem that Williams modified. Gardner mentioned the original monkey problem but didn’t do into much detail as he did with Williams modification of +1 coconuts to the monkey.

In More Coconuts by S. King, he related each step with solving the solution to different steps in the story. He integrated numbers into his writing which I liked because in this method he broke it down into parts and showed the numbers that represent each step instead of having a giant solve with numbers everywhere and context that you have to read after. I feel like it would have been better for Gardener to take this approach rather than separating the numbers and the words into separate parts.