# 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 1 (Stars and Bars)

Suppose that $P(x)=ax^3+bx^2+cx+d.$ This problem is equivalent to counting the ordered quadruples $(a,b,c,d),$ where all of $a,b,c,$ and $d$ are integers from $0$ through $9$ such that $$P(-1)=-a+b-c+d=-9.$$ Let $a'=9-a$ and $c'=9-c.$ Note that both of $a'$ and $c'$ are integers from $0$ through $9.$ Moreover, the ordered quadruples $(a,b,c,d)$ and the ordered quadruples $(a',b,c',d)$ have one-to-one correspondence.

We rewrite the given equation as $(9-a)+b+(9-c)+d=9,$ or $$a'+b+c'+d=9.$$ By the stars and bars argument, there are $\binom{9+4-1}{4-1}=\boxed{\textbf{(D) } 220}$ ordered quadruples $(a',b,c',d).$

~pieater314159 ~MRENTHUSIASM

## Solution 2 (Casework)

Suppose that $P(x)=ax^3+bx^2+cx+d.$ This problem is equivalent to counting the ordered quadruples $(a,b,c,d),$ where all of $a,b,c,$ and $d$ are integers from $0$ through $9$ such that $P(-1)=-a+b-c+d=-9,$ which rearranges to $$b+d+9=a+c.$$ Note that $b+d+9$ is an integer from $9$ through $27,$ ........................ So, ......................... We construct the following table: $$\begin{array}{c||c|c|c|c|c||c} & & & & & & \\ [-2.5ex] \textbf{Row} & \boldsymbol{x_2} & \boldsymbol{x_3} & \boldsymbol{x_4} & \boldsymbol{x_5} & \boldsymbol{x_2x_3x_4 + x_3x_4x_5} & \textbf{Valid?} \\ [0.5ex] \hline & & & & & & \\ [-2ex] 1 & 1 & 1 & 2 & 2 & 0 & \checkmark \\ 2 & 1 & 2 & 1 & 2 & 0 & \checkmark \\ 3 & 1 & 2 & 2 & 1 & 2 & \\ 4 & 2 & 1 & 1 & 2 & 1 & \\ 5 & 2 & 1 & 2 & 1 & 0 & \checkmark \\ 6 & 2 & 2 & 1 & 1 & 0 & \checkmark \end{array}$$

~BJHHar ~MRENTHUSIASM

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