Difference between revisions of "2018 AMC 12B Problems/Problem 22"
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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 | <cmath>-9=b+d-a-c.</cmath>If we let <math>b=-9-b', d=-9-d'</math> then we have | ||
<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> | <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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+ | Note: I think the equation should have been changed to: | ||
+ | <cmath>a'+ c' +b+d = 9</cmath> | ||
+ | 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 04:51, 25 January 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 The number of solutions to this equation is simply by stars and bars, so our answer is
Note: I think the equation should have been changed to: where and . This way all four variables are within 0 and 9.
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.