2011 IMO Problems/Problem 1
Given any set of four distinct positive integers, we denote the sum by . Let denote the number of pairs with for which divides . Find all sets of four distinct positive integers which achieve the largest possible value of .
Author: Fernando Campos, Mexico
Firstly, if we order , we see , so isn't a couple that satisfies the conditions of the problem. Also, , so again isn't a good couple. We have in total 6 couples. So .
We now find all sets with . If and are both good couples, and , we have . So WLOG with and . It's easy to see and since are bad, all couples containing must be good. Obviously and are good (). So we have and .
Using the second equation, we see that if , , for some a positive integer.
So now we use the first equation to get , for a natural .
Finally, we obtain 1, 2 or 4. We divide in cases:
CASE I: . So and . But 3, 4,5 or 6. implies , impossible. when . We easily see and , impossible since . When , , and we get .Uf , and we get .
CASE II and III:2, 4. Left to the reader.
ANSWER: ,, for any positive integer .
(Note: The above solution looks generally correct, but the actual answer should be ,. You can check that doesn't actually work. -Someone who didn't write up the above solution but solved the problem in a similar way)
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