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...") |
|||
Line 16: | Line 16: | ||
{{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}} | ||
+ | |||
+ | == 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 16:48, 7 June 2018
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 The number of solutions to this equation is simply by stars and bars, so our answer is
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.
Solution 2
Suppose our polynomial is equal to Then we are given that Then the polynomials , , also meet the criteria. So the number of solutions must be divisible by 4. So the answer must be