coconut

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 left after the previous man removed his (i.e., a_1 for the first man, a_2, for the second, ..., a_n for the last) and gives m coconuts (specified in the problem to be the same number for each man) which do not divide equally to a monkey. When all n men have so divided, they divide the remaining coconuts n ways (i.e., taking an additional a coconuts each), and give the m coconuts which are left over to the monkey. If m is the same at each division, then how many coconuts N were there originally? The solution is equivalent to solving the n + 1 Diophantine equations N | = | n a_1 + m N - a_1 - m | = | n a_2 + m N - a_1 - a_2 - 2m | = | n a_3 + m | ⋮ | N - a_1 - a_2 - a_3 - ... - a_n - n m | = | n a + m, which can be rewritten as N | = | n a_1 + m (n - 1) a_1 | = | n a_2 + m (n - 1) a_2 | = | n a_3 + m | ⋮ | (n - 1) a_(n - 1) | = | n a_n + m (n - 1) a_n | = | n a + m. Since there are n + 1 equations in the n + 2 unknowns a_1, a_2, ..., a_n, a, and N, 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 N = k n^(n + 1) - m(n - 1), where k is an arbitrary integer. 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 1024N = 15625a + 11529, where a is the number given to each man in the last division. The smallest positive solution in this case is N = 15621 coconuts, corresponding to k = 1 and a = 1023; 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 | | 15621 3124 | 1 | 12496 2499 | 1 | 9996 1999 | 1 | 7996 1599 | 1 | 6396 1279 | 1 | 5116 5×1023 | 1 | 0 If no coconuts are left for the monkey after the final n-way division, then the original number of coconuts is {(1 + n k) n^n - (n - 1) | n odd (n - 1 + n k) n^n - (n - 1) | n even. auto right match The smallest positive solution for case n = 5 and m = 1 is N = 3121 coconuts, corresponding to k = 0 and 1020 coconuts in the final division. The following table shows how these coconuts are divided. removed | given to monkey | left | 3121 | 624 | 1 | 2496 499 | 1 | 1996 399 | 1 | 1596 319 | 1 | 1276 255 | 1 | 1020 5×204 | 0 | 0 A different version of the problem having a solution of 79 coconuts is considered by Pappas.

Diophantine equation | Pell equation

coconut | macaques, mangabeys, baboons, guenons and monkeys