**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.