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 have $b+d-a-c=-9$ when $x=-1$ . So the number of solutions must be divisible by 4. So the answer must be $\boxed{\textbf{D}.}$

2018 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 21	Followed by Problem 23
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All AMC 12 Problems and Solutions

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2018 AMC 12B Problems/Problem 22

Contents

Problem

Solution

Solution 2

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