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 $-a=a'-9, -c=c'-9$ then we have \[9=a'+c'+b+d.\] This way all four variables are within 0 and 9. 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 $\boxed{\textbf{D}.}$


Solution 3

As before, $-9=-A+B-C+D$. This is $9=(A+B)-(C+D)$. Rephrased, how many two sums of integers from 0-9 have a difference of 9. Make a chart of pairs between these two sets: \[{0,1,2,3,4,5,6,7,8,9}\] \[{9,10,11,12,13,14,15,16,17,18}\] Observe how there is one way to sum 2 numbers to 0 and two ways to 1, 3 ways to 2, and so on. At 9, there are 10 possible ways. Recall that only integers between 0-9 are valid. Now observe how there is 1 way to to sum to 18 in this fashion (9+9), 2 ways to sum to 17, and so forth again (to optionally prove that this pattern holds, apply stars and bars up to 9 and notice the symmetry).

The answer then is the number of ways to write each component of each pair. This is $(1*10)+(2*9)+(3*8) ...$ or, since it's symmetrical between sum of 4 and 5, $[2\sum\limits_{i=1}^{5} i(11-i)]$. Use summation rules to finally get $\boxed{\textbf{D)}220}$.

~BJHHar

See Also

2018 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 21
Followed by
Problem 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. AMC logo.png

Invalid username
Login to AoPS