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Difference between revisions of "2016 AIME II Problems/Problem 6"

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Note: this method doesn't work for every product of polynomials. Example: The sum of the absolute values of the coefficients of <math>K(x) = (2x^2 - 6x - 5)(10x - 1)</math> is <math>131</math> but when you find the sum of the coefficients of <math>K'(x)</math> which is <math>K'(1)</math>, then you get <math>143</math>. - whatRthose
 
Note: this method doesn't work for every product of polynomials. Example: The sum of the absolute values of the coefficients of <math>K(x) = (2x^2 - 6x - 5)(10x - 1)</math> is <math>131</math> but when you find the sum of the coefficients of <math>K'(x)</math> which is <math>K'(1)</math>, then you get <math>143</math>. - whatRthose
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==Solution 3 (risky)==
 +
Multiply <math>P(x)P(x^3)</math> and notice that the odd degree terms have a negative coefficient. Observing that this is probably true for all polynomials like this (including <math>P(x)P(x^3)P(x^5)P(x^7)P(x^9)</math>), we plug in <math>-1</math> to get <math>\frac{243}{32} \implies \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 13:36, 8 March 2020

Problem

For polynomial $P(x)=1-\dfrac{1}{3}x+\dfrac{1}{6}x^{2}$, define $Q(x)=P(x)P(x^{3})P(x^{5})P(x^{7})P(x^{9})=\sum_{i=0}^{50} a_ix^{i}$. Then $\sum_{i=0}^{50} |a_i|=\dfrac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

Solution 1

Note that all the coefficients of odd-powered terms is an odd number of odd degree terms multiplied together, and all coefficients of even-powered terms 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 $Q(-1)=P(-1)^{5}=\left( \dfrac{3}{2}\right)^{5}=\dfrac{243}{32}$, so the desired answer is $243+32=\boxed{275}$.

Solution by Shaddoll

Solution 2

We are looking for the sum of the absolute values of the coefficients of $Q(x)$. By defining $P'(x) = 1 + \frac{1}{3}x+\frac{1}{6}x^2$, and defining $Q'(x) = P'(x)P'(x^3)P'(x^5)P'(x^7)P'(x^9)$, we have made it so that all coefficients in $Q'(x)$ are just the positive/absolute values of the coefficients of $Q(x)$. .


To find the sum of the absolute values of the coefficients of $Q(x)$, we can just take the sum of the coefficients of $Q'(x)$. This sum is equal to \[Q'(1) = P'(1)P'(1)P'(1)P'(1)P'(1) = \left(1+\frac{1}{3}+\frac{1}{6}\right)^5 = \frac{243}{32},\]

so our answer is $243+32 = \boxed{275}$.


Solution by ExuberantGPACN

Note: this method doesn't work for every product of polynomials. Example: The sum of the absolute values of the coefficients of $K(x) = (2x^2 - 6x - 5)(10x - 1)$ is $131$ but when you find the sum of the coefficients of $K'(x)$ which is $K'(1)$, then you get $143$. - whatRthose

Solution 3 (risky)

Multiply $P(x)P(x^3)$ and notice that the odd degree terms have a negative coefficient. Observing that this is probably true for all polynomials like this (including $P(x)P(x^3)P(x^5)P(x^7)P(x^9)$), we plug in $-1$ to get $\frac{243}{32} \implies \boxed{275}$.

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
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