 If there
were no remainders, given there are 6 divisions of 5, 56=15,625 coconuts must be in the pile; on
the 6th and last division, each sailor receives 1024 coconuts.No smaller positive number
will result in all 6 divisions coming out even.That means that in the problem as stated,
any multiple of 15,625 may be added to the pile, and it will satisfy the problem conditions.