Difference between revisions of "2016 AIME II Problems/Problem 6"
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Then <math>\sum_{i=0}^{50} |a_i|=\dfrac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>. | Then <math>\sum_{i=0}^{50} |a_i|=\dfrac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>. | ||
− | ==Solution== | + | ==Solution 1== |
Note that all the odd coefficients have an odd number of odd degree terms multiplied together, and all the even coefficients have an even number of odd degree terms multiplied together. Since every odd degree term is negative, and every even degree term is positive, the sum is just equal to <math>Q(-1)=P(-1)^{5}=\left( \dfrac{3}{2}\right)^{5}=\dfrac{243}{32}</math>, so the desired answer is <math>243+32=\boxed{275}</math>. | Note that all the odd coefficients have an odd number of odd degree terms multiplied together, and all the even coefficients have an even number of odd degree terms multiplied together. Since every odd degree term is negative, and every even degree term is positive, the sum is just equal to <math>Q(-1)=P(-1)^{5}=\left( \dfrac{3}{2}\right)^{5}=\dfrac{243}{32}</math>, so the desired answer is <math>243+32=\boxed{275}</math>. | ||
Solution by Shaddoll | Solution by Shaddoll | ||
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+ | ==Solution 2== | ||
+ | We are looking for the sum of the absolute values of the coefficients of <math>Q(x)</math>. By defining <math>P'(x) = 1 + \frac{1}{3}x+\frac{1}{6}x^2</math>, and defining <math>Q'(x) = P'(x)P'(x^3)P'(x^5)P'(x^7)P'(x^9)</math>, we have made it so that all coefficients in <math>Q'(x)</math> are just the positive/absolute values of the coefficients of <math>Q(x)</math>. . | ||
+ | |||
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+ | To find the sum of the absolute values of the coefficients of <math>Q(x)</math>, we can just take the sum of the coefficients of <math>Q'(x)</math>. This sum is equal to | ||
+ | <cmath>Q'(1) = P'(1)P'(1)P'(1)P'(1)P'(1) = \left(1+\frac{1}{3}+\frac{1}{6}\right)^5 = \frac{243}{32},</cmath> | ||
+ | |||
+ | so our answer is <math>243+32 = \boxed{275}</math>. | ||
== See also == | == See also == | ||
{{AIME box|year=2016|n=II|num-b=5|num-a=7}} | {{AIME box|year=2016|n=II|num-b=5|num-a=7}} |
Revision as of 12:04, 6 July 2016
For polynomial , define . Then , where and are relatively prime positive integers. Find .
Solution 1
Note that all the odd coefficients have an odd number of odd degree terms multiplied together, and all the even coefficients have an even number of odd degree terms multiplied together. Since every odd degree term is negative, and every even degree term is positive, the sum is just equal to , so the desired answer is .
Solution by Shaddoll
Solution 2
We are looking for the sum of the absolute values of the coefficients of . By defining , and defining , we have made it so that all coefficients in are just the positive/absolute values of the coefficients of . .
To find the sum of the absolute values of the coefficients of , we can just take the sum of the coefficients of . This sum is equal to
so our answer is .
See also
2016 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 5 |
Followed by Problem 7 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |