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
 |  |  |
(1)
|
 |  |  |
(2)
|
 |  |  |
(3)
|
 |  |  |
(4)
|
 |  |  |
(5)
|
which can be rewritten as
 |  |  |
(6)
|
 |  |  |
(7)
|
 |  |  |
(8)
|
 |  |  |
(9)
|
 |  |  |
(10)
|
 |  |  |
(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
 |
(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
 |
(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 |
 | ||
 | 1 |  |
 | 1 |  |
 | 1 |  |
 | 1 |  |
 | 1 |  |
 | 1 | 0 |
If no coconuts are left for the monkey after the final 
-way division (Williams 1926), then the original number of
coconuts is
 |
(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 |
 | ||
| 624 | 1 |  |
| 499 | 1 |  |
| 399 | 1 |  |
| 319 | 1 |  |
| 255 | 1 |  |
 | 0 | 0 |
A different version of the problem having a solution of 79 coconuts is considered by Pappas (1989).
." Amer. Math. Monthly 35, 47-48, 1928.Ogilvy,
C. S. and Anderson, J. T.