They published their solution and it’s kind of weak:
SOLUTION: You can solve this by considering two men instead of five, then three, then guessing. But the following argument is irresistible, once found.
There's an elegant "solution" to the puzzle if you allow negative numbers of coconuts(!). The original pile has -4 coconuts; when the first man tosses the monkey a coconut, the pile is down to -5, but when he "takes" 1/5 of this he is actually adding a coconut, restoring the pile to -4 coconuts. Continuing this way, come morning there are still -4 coconuts; the monkey takes one and the men split up the remaining -5.
It's not obvious that this observation does us any good, but let's consider what happens if there is no monkey; each man just takes 1/5 of the pile he encounters, and in the morning there's a multiple of 5 coconuts left that the men can split. Since each man has reduced the pile by the fraction 4/5, the original number of coconuts must have been a multiple of 56 (which shrinks to 45 x 5 by morning).
All we need to do now is add our two pseudo-solutions, by starting with 56 – 4 = 15,621 coconuts. Then the pile reduces successively to 4 x 55 – 4 coconuts, 42 x 54 – 4, 43 x 53 – 4, 44 x 52 – 4, and 45 x 5 – 4. When the monkey gets his morning coconut, we have 45 x 5 – 5 coconuts, a multiple of 5, for the men to split. This is best possible because we needed 55 x k – 4 coconuts to start with, just to have an integer number come morning, and to get 45 x k – 5 to be a multiple of 5 we needed k to be a multiple of 5.
I emailed my solution as more concise.
