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

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

Revision as of 18:59, 21 October 2018

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

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

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