# Difference between revisions of "2018 AMC 12B Problems/Problem 22"

## Problem

Consider polynomials $P(x)$ of degree at most $3$, each of whose coefficients is an element of $\{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\}$. How many such polynomials satisfy $P(-1) = -9$?

$\textbf{(A) } 110 \qquad \textbf{(B) } 143 \qquad \textbf{(C) } 165 \qquad \textbf{(D) } 220 \qquad \textbf{(E) } 286$

## Solution

Suppose our polynomial is equal to $$ax^3+bx^2+cx+d$$Then we are given that $$-9=b+d-a-c.$$If we let $b=-9-b', d=-9-d'$ then we have $$9=a+c+b'+d'.$$The number of solutions to this equation is simply $\binom{12}{3}=220$ by stars and bars, so our answer is $\boxed{\textbf{D}.}$

Note: I think the equation should have been changed to: $$a'+ c' +b+d = 9$$ where $a'=9-a$ and $c'=9-c$. This way all four variables are within 0 and 9.

## Solution 2

Suppose our polynomial is equal to $$ax^3+bx^2+cx+d$$Then we are given that $$9=b+d-a-c.$$Then the polynomials $$cx^3+bx^2+ax+d$$, $$ax^3+dx^2+cx+b,$$ $$cx^3+dx^2+ax+b$$also have $$b+d-a-c=-9$$ when $$x=-1.$$ So the number of solutions must be divisible by 4. So the answer must be $\boxed{\textbf{D}.}$