2016 AIME II Problems/Problem 6

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

2016 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 5	Followed by Problem 7
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2016 AIME II Problems/Problem 6

Solution 1

Solution 2

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