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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
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<cmath>-9=b+d-a-c.</cmath>If we let <math>b=-9-b', d=-9-d'</math> then we have
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<cmath>-9=b+d-a-c.</cmath>If we let <math>-a=a'-9, -c=c'-9</math> then we have
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<cmath>9=a+c+b'+d'.</cmath>The number of solutions to this equation is simply <math>\binom{12}{3}=220</math> by stars and bars, so our answer is <math>\boxed{\textbf{D}.}</math>
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<cmath>9=a'+c'+b+d.</cmath> This way all four variables are within 0 and 9. The number of solutions to this equation is simply <math>\binom{12}{3}=220</math> by stars and bars, so our answer is <math>\boxed{\textbf{D}.}</math>
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Note: I think the equation should have been changed to:
 
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<cmath>a'+ c' +b+d = 9</cmath>
 
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where <math>a'=9-a</math> and <math>c'=9-c</math>. This way all four variables are within 0 and 9.
 
  
 
== Solution 2 ==
 
== Solution 2 ==

Revision as of 14:46, 18 July 2019

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

See Also

2018 AMC 12B (Problems • Answer Key • Resources)
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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