Coconut Problem Posted by Mattw:
If they'd JUST gathered four more coconuts, then the pile would have split evenly, and four more coconuts would pass to the next stage. Five times four extra coconuts would allow the the pile to reduce by exactly 4/5 and also it would split evenly in five shares at the end. If T was the original pile size, then (T+4)*(4/5)^5*(1/5) would be the resulting coconut share, an integer. Thus T+4 must be a positive multiple of 5^6 = 15625. The smallest solution for T is 15621. Generally T = 15625*n - 4 works for any positive integer n. (original solution)

By the way, regarding the 12-coins Problem: (Nyah-Nyah, nobody got it yet.) There are only 2 platters, not three. Moreover, you should also determine whether the result is lighter or heavier.

I just figured out that on the front page the Recent Diary Entries link shows them together in order. That is swell.