Coconuts!

[math7839]

1600 coconuts are distributed among 100 monkeys. Some monkeys may receive zero coconuts.
Prove that no matter how the distribution is done, there are always at least four monkeys that receive the same number of coconuts.
Can you figure it out?

 

[mmm4444bot]

If we assume that each coconut must be distributed intact, here are some lines of reasoning.



Distribute one coconut to the first monkey

Distribute two coconuts to the second monkey

Distribute three coconuts to the third monkey
.
.
.
Distribute 56 coconuts to the 56th monkey

Since 56*57/2 = 1596, we have (thus far) distributed 1,596 coconuts among 56 monkeys, with each of those monkeys possessing a different number of coconuts.

Hence, four coconuts remain to be distributed among the 44 monkeys who each have zero coconuts.

Clearly, a whole bunch 'o monkeys will end up with the same number of coconuts (zero).



Distribute one coconut to monkeys 1, 2, and 3

Distribute two coconuts to monkeys 4, 5, and 6

Distribute three coconuts to monkeys 7, 8, and 9
.
.
.
Distribute 32 coconuts to monkeys 94, 95, and 96

Since 3(32*33)/2 = 1584, we have (this time) distributed 1,584 coconuts among 96 monkeys, with no more than three of those monkeys possessing the same number of coconuts.

Hence, 16 coconuts remain to be distributed among the four monkeys who each have zero coconuts.

Clearly, giving any number of the remaining 16 coconuts to any of these four monkeys will result in four monkeys (overall) having the same number of coconuts.



Do these lines of reasoning help you to write a formal proof?

(NOTE: I did not double-check my work.)