Difference between revisions of "2020 AIME I Problems/Problem 11"
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Revision as of 12:48, 28 March 2020
For integers and let and Find the number of ordered triples of integers with absolute values not exceeding for which there is an integer such that
Solution 1 (Strategic Casework)
Either or not. If it is, note that Vieta's forces . Then, can be anything. However, can also be anything, as we can set the root of (not equal to ) to any integer, producing a possible integer value of . Therefore there are in this case. If it isn't, then are the roots of . This means by Vieta's, that:
Solving these inequalities while considering that to prevent , we obtain possible tuples and adding gives . ~awang11
Solution 2 (Bash)
Define . Since , we know . Plugging in into , we get . Setting , . Simplifying and cancelling terms,
Therefore, either or . The first case is easy: and there are tuples in that case. In the second case, we simply perform casework on even values of , to get tuples, subtracting the tuples in both cases we get .
In case anyone is confused by this (as I initially was). Say . This does not mean that g has a double root of , ONLY that is one of the roots of g. ~First
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