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>If we let <math> | + | <cmath>-9=b+d-a-c.</cmath>If we let <math>-a=a'-9, -c=c'-9</math> then we have |
− | <cmath>9=a+c+b | + | <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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== Solution 2 == | == Solution 2 == |
Revision as of 14:46, 18 July 2019
Contents
Problem
Consider polynomials of degree at most , each of whose coefficients is an element of . How many such polynomials satisfy ?
Solution
Suppose our polynomial is equal to Then we are given that If we let then we have This way all four variables are within 0 and 9. The number of solutions to this equation is simply by stars and bars, so our answer is
Solution 2
Suppose our polynomial is equal to Then we are given that Then the polynomials , also have when So the number of solutions must be divisible by 4. So the answer must be
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.