Difference between revisions of "2011 IMO Problems/Problem 1"
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CASE II and III:<math>k_1=</math>2, 4. Left to the reader. | CASE II and III:<math>k_1=</math>2, 4. Left to the reader. | ||
− | ANSWER: <math>\{11a, a, 5a, 7a\}</math>,<math>\{a, | + | ANSWER: <math>\{11a, a, 5a, 7a\}</math>,<math> \{a, 11a, 19a, 29a\} </math>, for any positive integer <math>a</math>. |
(Note: The above solution looks generally correct, but the actual answer should be <math>\{11a, a, 5a, 7a\}</math>,<math>\{a, 11a, 19a, 29a\}</math>. You can check that <math>\{a, 14a, 6a, 9a\}</math> doesn't actually work. -Someone who didn't write up the above solution but solved the problem in a similar way) | (Note: The above solution looks generally correct, but the actual answer should be <math>\{11a, a, 5a, 7a\}</math>,<math>\{a, 11a, 19a, 29a\}</math>. You can check that <math>\{a, 14a, 6a, 9a\}</math> doesn't actually work. -Someone who didn't write up the above solution but solved the problem in a similar way) |
Revision as of 22:07, 19 June 2014
Problem
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
Solution
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)
See Also
2011 IMO (Problems) • Resources | ||
Preceded by First question |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 2 |
All IMO Problems and Solutions |