Bob has 10 pockets and 44 silver dollars. He wants to put his dollars into his pockets so distributed that each pocket contains a different number of dollars.
- Can he do that?
- Generalize the problem, considering p pockets and n dollars. The problem is the most interesting when n = (p + 1)(p - 2)/2. Why?
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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
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n = p(p - 1)/2 - 1 = (p² - p - 2)/2 = (p + 1)(p - 2)/2.
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However, if "putting" $0 into a pocket is not a legitimate action, then the least amount required is given by
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1 + 2 + ... + p = p(p + 1)/2.
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The borderline number is one less:
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n = p(p + 1)/2 - 1 = (p² + p - 2)/2 = (p - 1)(p + 2)/2.
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Reference
- G. Polya, J. Kilpatrick, The Stanford Mathematics Problem Book, Dover, 2009, #53.1
Copyright © 1996-2009 Alexander Bogomolny
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