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Consider the first 99 monkeys. The least number of bananas that does not give 4 monkeys the same amount, is distributed such that monkey 1,2,3 get nothing, monkey 4,5,6 get one banana, monkey 7,8,9 get two and so forth. We end up distributing 3*33*32/2 = 1584 bananas. That leaves 16 bananas for the monkey #100, an amount that has already been given to 3 others. If we only had 17 bananas more we could give 33 bananas to #100 and we'd be fine.
See also Old Original Oskar!'s posting (1617 reasons...) |
