Question

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

Answer

Answer: 275

Concepts:polynomial substitution

Difficulty rating: 2400

Solution:

Every substituted power $x, x^3, x^5, x^7, x^9$ is odd, so $Q(-x)$ $= P(-x)P(-x^3)P(-x^5)$ $\cdot P(-x^7)P(-x^9).$ Since $P(-x) = 1 + \frac{1}{3}x + \frac{1}{6}x^2$ has only nonnegative coefficients, so does each factor $P(-x^k),$ and hence so does the product $Q(-x).$ The coefficient of $x^i$ in $Q(-x)$ is $(-1)^i a_i,$ so $|a_i| = (-1)^i a_i.$

Therefore $\begin{aligned} \sum_{i=0}^{50} |a_i| &= Q(-1) = P(-1)^5 \\ &= \left(1 + \frac{1}{3} + \frac{1}{6}\right)^5 \\ &= \left(\frac{3}{2}\right)^5 = \frac{243}{32}, \end{aligned}$ and $m + n = 243 + 32 = 275.$

2016 AIME II Problem 6

Answer: 275

Difficulty rating: 2400

Solution:

Problem 6 in Other Years