# 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 $\boxed{\textbf{D}.}$

## 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 meet the criteria. So the number of solutions must be divisible by 4. So the answer must be $\boxed{\textbf{D}.}$

## See Also

 2018 AMC 12B (Problems • Answer Key • Resources) Preceded byProblem 21 Followed byProblem 23 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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