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Monkey and Coconut Problem

A Diophantine problem (i.e., one whose solution must be given in terms of Integers) which seeks a solution to the following problem. Given $n$ men and a pile of coconuts, each man in sequence takes $(1/n)$th of the coconuts and gives the $m$ coconuts which do not divide equally to a monkey. When all $n$ men have so divided, they divide the remaining coconuts five ways, and give the $m$ coconuts which are left-over to the monkey. How many coconuts $N$ were there originally? The solution is equivalent to solving the $n+1$ Diophantine Equations

$\displaystyle N$ $\textstyle =$ $\displaystyle nA+m$  
$\displaystyle (n-1)A$ $\textstyle =$ $\displaystyle nB+m$  
$\displaystyle (n-1)B$ $\textstyle =$ $\displaystyle nC+m$  
  $\textstyle \vdots$    
$\displaystyle (n-1)X$ $\textstyle =$ $\displaystyle nY+m$  
$\displaystyle (n-1)Y$ $\textstyle =$ $\displaystyle nZ+m,$  

and is given by


where $k$ is an arbitrary Integer (Gardner 1961).

For the particular case of $n=5$ men and $m=1$ left over coconuts, the 6 equations can be combined into the single Diophantine Equation


where $F$ is the number given to each man in the last division. The smallest Positive solution in this case is $N=15,621$ coconuts, corresponding to $k=1$ and $F=1,023$ (Gardner 1961). The following table shows how this rather large number of coconuts is divided under the scheme described above.

Removed Given to Monkey Left
3,124 1 12,496
2,499 1 9,996
1,999 1 7,996
1,599 1 6,396
1,279 1 5,116
$5\times 1023$ 1 0

If no coconuts are left for the monkey after the final $n$-way division (Williams 1926), then the original number of coconuts is

(1+nk)n^n-(n-1) & $n$\ odd\cr
(n-1+nk)n^n-(n-1) & $n$\ even.\cr}

The smallest Positive solution for case $n=5$ and $m=1$ is $N=3,121$ coconuts, corresponding to $k=1$ and 1,020 coconuts in the final division (Gardner 1961). The following table shows how these coconuts are divided.

Removed Given to Monkey Left
624 1 2,496
499 1 1,996
399 1 1,596
319 1 1,276
255 1 1,020
$5\times 204$ 0 0

A different version of the problem having a solution of 79 coconuts is considered by Pappas (1989).

See also Diophantine Equation--Linear, Pell Equation


Anning, N. ``Monkeys and Coconuts.'' Math. Teacher 54, 560-562, 1951.

Bowden, J. ``The Problem of the Dishonest Men, the Monkeys, and the Coconuts.'' In Special Topics in Theoretical Arithmetic. Lancaster, PA: Lancaster Press, pp. 203-212, 1936.

Gardner, M. ``The Monkey and the Coconuts.'' Ch. 9 in The Second Scientific American Book of Puzzles & Diversions: A New Selection. New York: Simon and Schuster, 1961.

Kirchner, R. B. ``The Generalized Coconut Problem.'' Amer. Math. Monthly 67, 516-519, 1960.

Moritz, R. E. ``Solution to Problem 3,242.'' Amer. Math. Monthly 35, 47-48, 1928.

Ogilvy, C. S. and Anderson, J. T. Excursions in Number Theory. New York: Dover, pp. 52-54, 1988.

Olds, C. D. Continued Fractions. New York: Random House, pp. 48-50, 1963.

Pappas, T. ``The Monkey and the Coconuts.'' The Joy of Mathematics. San Carlos, CA: Wide World Publ./Tetra, pp. 226-227 and 234, 1989.

Williams, B. A. ``Coconuts.'' The Saturday Evening Post, Oct. 9, 1926.

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© 1996-9 Eric W. Weisstein