2016 AIME II Problem 6

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

The coefficients of

P(-x)

are all positive, and

Q(-x)

is a product of such polynomials

So

\sum |a_i|

is just

Q(-1) = P(-1)^5

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.

Problem 6 in other years