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

(Created page with "== Problem == Consider polynomials <math>P(x)</math> of degree at most <math>3</math>, each of whose coefficients is an element of <math>\{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\}</mat...")
 
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{{AMC12 box|year=2018|ab=B|num-b=21|num-a=23}}
 
{{AMC12 box|year=2018|ab=B|num-b=21|num-a=23}}
 
{{MAA Notice}}
 
{{MAA Notice}}
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== Solution 2 ==
 +
Suppose our polynomial is equal to
 +
<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 meet the criteria. So the number of solutions must be divisible by 4. So the answer must be <math>\boxed{\textbf{D}.}</math>

Revision as of 17:48, 7 June 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 $\boxed{\textbf{D}.}$

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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 meet the criteria. So the number of solutions must be divisible by 4. So the answer must be $\boxed{\textbf{D}.}$

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