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Date: 19-12-2020
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Date: 20-11-2019
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Date: 22-5-2020
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A Diophantine problem (i.e., one whose solution must be given in terms of integers) which seeks a solution to the following problem. Given men and a pile of coconuts, each man in sequence takes
th of the coconuts left after the previous man removed his (i.e.,
for the first man,
, for the second, ...,
for the last) and gives
coconuts (specified in the problem to be the same number for each man) which do not divide equally to a monkey. When all
men have so divided, they divide the remaining coconuts
ways (i.e., taking an additional
coconuts each), and give the
coconuts which are left over to the monkey. If
is the same at each division, then how many coconuts
were there originally? The solution is equivalent to solving the
Diophantine equations
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(1) |
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(2) |
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(3) |
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(4) |
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(5) |
which can be rewritten as
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(6) |
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(7) |
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(8) |
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(9) |
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(10) |
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(11) |
Since there are equations in the
unknowns
,
, ...,
,
, and
, the solutions span a one-dimensional space (i.e., there is an infinite family of solution parameterized by a single value). The solution to these equations can be given by
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(12) |
where is an arbitrary integer (Gardner 1961).
For the particular case of men and
left over coconuts, the 6 equations can be combined into the single Diophantine equation
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(13) |
where is the number given to each man in the last division. The smallest positive solution in this case is
coconuts, corresponding to
and
; 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 |
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||
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1 | ![]() |
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1 | ![]() |
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1 | ![]() |
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1 | ![]() |
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1 | ![]() |
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1 | 0 |
If no coconuts are left for the monkey after the final -way division (Williams 1926), then the original number of coconuts is
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(14) |
The smallest positive solution for case and
is
coconuts, corresponding to
and
coconuts in the final division (Gardner 1961). The following table shows how these coconuts are divided.
removed | given to monkey | left |
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||
624 | 1 | ![]() |
499 | 1 | ![]() |
399 | 1 | ![]() |
319 | 1 | ![]() |
255 | 1 | ![]() |
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0 | 0 |
A different version of the problem having a solution of 79 coconuts is considered by Pappas (1989).
REFERENCES:
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, pp. 104-111, 1961.
Kirchner, R. B. "The Generalized Coconut Problem." Amer. Math. Monthly 67, 516-519, 1960.
Moritz, R. E. "Solution to Problem ." 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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