h2g2 The Hitchhiker's Guide to the Galaxy: Earth Edition

There's a slightly easier solution to this puzzle. Taking the equations presented in the solution:

5A + 1 = N
5B + 1 = 4A
5C + 1 = 4B
5D + 1 = 4C
5E + 1 = 4D
5F + 1 = 4E

Add 4 to each equation, giving:

5A + 5 = N + 4 => 5(A + 1) = N + 4
5B + 5 = 4A + 4 => 5(B + 1) = 4(A + 1)
5C + 5 = 4B + 4 => 5(C + 1) = 4(B + 1)
5D + 5 = 4C + 4 => 5(D + 1) = 4(C + 1)
5E + 5 = 4D + 4 => 5(E + 1) = 4(D + 1)
5F + 5 = 4E + 4 => 5(F + 1) = 4(E + 1)

Therefore N + 4 = 5 * (5/4)^5 * (F + 1), and so
N = (5^6/4^5)(F + 1) - 4.

Since 5 and 4 are coprime, 5^6/4^5 = 15625/1024 is a fraction in its lowest terms. Hence the only integer solutions of the above equation are where F + 1 is a multiple of 4^5.

So the general solution is N = 15625r - 4, where r is a positive integer, giving a smallest solution of 15621 coconuts in the original pile.