2020 AIME I Problems/Problem 11
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Problem
== Solution ==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
See Also
2020 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
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