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

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so our answer is <math>243+32 = \boxed{275}</math>.
 
so our answer is <math>243+32 = \boxed{275}</math>.
  
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~ExuberantGPACN
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Solution by ExuberantGPACN
  
 
== 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 12:06, 6 July 2016

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 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 $Q(-1)=P(-1)^{5}=\left( \dfrac{3}{2}\right)^{5}=\dfrac{243}{32}$, 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

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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