Difference between revisions of "2020 AIME I Problems/Problem 11"
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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 | 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 | ||
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+ | == Solution 2 (Bash) == | ||
+ | Define <math>h(x)=x^2+cx</math>. Since <math>g(f(2))=g(f(4))=0</math>, we know <math>h(f(2))=h(f(4))=-d</math>. Plugging in <math>f(x)</math> into <math>h(x)</math>, we get <math>h(f(x))=x^4+2ax^3+(2b+a^2+c)x^2+(2ab+ac)x+(b^2+bc)</math>. Setting <math>h(f(2))=h(f(4))</math>, <cmath>16+16a+8b+4a^2+4ab+b^2+4c+2ac+bc=256+128a+32b+16a^2+8ab+b^2+16c+4ac+bc</cmath>. Simplifying and cancelling terms, <cmath>240+112a+24b+12a^2+4ab+12c+2ac=0</cmath> <cmath>120+56a+12b+6a^2+2ab+6c+ac=0</cmath> <cmath>6a^2+2ab+ac+56a+12b+6c+120=0</cmath> <cmath>6a^2+2ab+ac+20a+36a+12b+6c+120=0</cmath> <cmath>a(6a+2b+c+20)+6(6a+2b+c+20)=0</cmath> <cmath>(a+6)(6a+2b+c+20)=0</cmath> | ||
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+ | Therefore, either <math>a+6=0</math> or <math>6a+2b+c=-20</math>. The first case is easy: <math>a=-6</math> and there are <math>441</math> tuples in that case. In the second case, we simply perform casework on even values of <math>c</math>, to get <math>77</math> tuples, subtracting the <math>8</math> tuples in both cases we get <math>441+77-8=\boxed{510}</math>. | ||
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+ | -EZmath2006 | ||
==See Also== | ==See Also== |
Revision as of 21:45, 12 March 2020
Contents
Problem
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
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 .
-EZmath2006
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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