Difference between revisions of "2020 AIME I Problems/Problem 11"
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== Problem == | == Problem == | ||
− | == Solution == | + | == Solution ==Either <math>f(2)=f(4)</math> or not. If it is, note that Vieta's forces <math>a = -6</math>. Then, <math>b</math> can be anything. However, <math>c</math> can also be anything, as we can set the root of <math>g</math> (not equal to <math>f(2) = f(4)</math>) to any integer, producing a possible integer value of <math>d</math>. Therefore there are <math>21^2 = 441</math> in this case. If it isn't, then <math>f(2),f(4)</math> are the roots of <math>g</math>. This means by Vieta's, that: |
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+ | <cmath>f(2)+f(4) = -c \in [-10,10]</cmath> | ||
+ | <cmath>20 + 6a + 2b \in [-10,10]</cmath> | ||
+ | <cmath>3a + b \in [-15,5].</cmath> | ||
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+ | Solving these inequalities while considering that <math>a \neq -6</math> to prevent <math>f(2) = f(4)</math>, we obtain <math>69</math> possible tuples and adding gives <math>441+69=\boxed{510}</math>. ~awang11 | ||
==See Also== | ==See Also== |
Revision as of 17:23, 12 March 2020
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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 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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