Dollars in Pockets

There is a tacit assumption that Bob can put 0 dollars into a pocket which is common in mathematics, although in reality one is allowed to question whether it is proper to use a verb ("put" in this case) where no action has been performed.

So, assuming 0 is the least possible amount Bob can put into a pocket, the other small amounts would be

If ten positive integers add up to less than 45, there is no way to place different amounts into 10 pockets. Some two will have to be equal.

Where does the quantity n = (p + 1)(p - 2)/2 come from? For the number p of the pockets, the least amount that satisfies the requirement of the problem is the sum 0 + 1 + 2 + ... + (p-1) = p(p - 1)/2. One less and the problem becomes unsolvable. The borderline n is then

However, if "putting" $0 into a pocket is not a legitimate action, then the least amount required is given by

The borderline number is one less:

Reference

1. G. Polya, J. Kilpatrick, The Stanford Mathematics Problem Book, Dover, 2009, #53.1

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