Five men were stuck in a desert island. The island was plentifully supplied with coconuts, fortunately, and they spent the first day gathering the coconuts, heaping them in a great pile.

That night, one of the men was sleepless at the thought that everyone would fight over the coconuts at daybreak. So he arose, crept to the pile, and counted out the coconuts. Except for one coconut, it was possible to divide the pile into five equal portions. He gave the extra coconut to the monkey (who was the only other living being on the island besides those five men), buried his share of the coconuts, then went back to a sound sleep leaving the rest again in a pile.

Then another man got up and stole to the pile with the same object in mind. Again there was one coconut left over. The second man gave the spare coconut to the monkey, buried (what he thought was) his fair share, and went back to sleep, again leaving the remaining coconuts in one pile.

One by one, the remaining men got up and did the same and everytime the monkey got his share of one coconut.

In the morning, the men divided whatever coconuts remained into five equal piles, each taking his share and saying nothing about the night just past.

What is the minimum number of coconuts the five men assembled on the first day, if in the Final Division:

(a)There was again one coconut left over and the men (as usual) gave it to the monkey.

(b) There was no coconut left over for the monkey.

(In reply to re: Comment on official solution by Ravi Raja)

Ravi, I realise there are methods to solve Diophantine equations.  My point was that, having written out in great detail the equations connecting x and A, A and B, B and C, C and D, D and E, and E and y, you then completely omit the (non-trivial) solution method for the resulting Diophantine equation.  Anyone who has not met this type of equation before will be totally mystified as to how you obtained the answer!

In any case, there is an easier solution that does not use Diophantine equations.

Consider part (b), where, using your notation, we have the following equations:

4x - 4 = 5A
4A - 4 = 5B
4B - 4 = 5C
4C - 4 = 5D
4D - 4 = 5E
E = 5y

Adding 16 to both sides of the first five equations, we get

4(x + 4) = 5(A + 4)
4(A + 4) = 5(B + 4)
4(B + 4) = 5(C + 4)
4(C + 4) = 5(D + 4)
4(D + 4) = 5(E + 4)

Hence x + 4 = (5/4)⁵(E + 4) = (5/4)⁵(5y + 4)

Since (5/4)⁵ is a fraction in its lowest terms, the only integer solutions are where 5y + 4 is a multiple of 4⁵.  The smallest solution is where 5y + 4 = 4⁵, in which case x + 4 = 5⁵ = 3125.

Part (a) may be solved similarly.

Posted by Nick Hobson on 2005-06-01 21:41:10