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# Difference between revisions of "2020 AIME I Problems/Problem 11"

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## Solution

Either $f(2)=f(4)$ or not. If it is, note that Vieta's forces $a = -6$. Then, $b$ can be anything. However, $c$ can also be anything, as we can set the root of $g$ (not equal to $f(2) = f(4)$) to any integer, producing a possible integer value of $d$. Therefore there are $21^2 = 441$ in this case. If it isn't, then $f(2),f(4)$ are the roots of $g$. This means by Vieta's, that:

$$f(2)+f(4) = -c \in [-10,10]$$ $$20 + 6a + 2b \in [-10,10]$$ $$3a + b \in [-15,5].$$

Solving these inequalities while considering that $a \neq -6$ to prevent $f(2) = f(4)$, we obtain $69$ possible tuples and adding gives $441+69=\boxed{510}$. ~awang11

## See Also

 2020 AIME I (Problems • Answer Key • Resources) Preceded byProblem 10 Followed byProblem 12 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 All AIME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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