Difference between revisions of "2018 AMC 12B Problems/Problem 22"
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As before, <math>-9=-A+B-C+D</math>. This is <math>9=(A+B)-(C+D)</math>. Rephrased, how many two sums of integers from 0-9 have a difference of 9. Make a chart of pairs between these two sets: | As before, <math>-9=-A+B-C+D</math>. This is <math>9=(A+B)-(C+D)</math>. Rephrased, how many two sums of integers from 0-9 have a difference of 9. Make a chart of pairs between these two sets: | ||
<cmath>{0,1,2,3,4,5,6,7,8,9}</cmath> | <cmath>{0,1,2,3,4,5,6,7,8,9}</cmath> |
Latest revision as of 23:42, 1 August 2020
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
Solution 3
As before, . This is . Rephrased, how many two sums of integers from 0-9 have a difference of 9. Make a chart of pairs between these two sets: Observe how there is one way to sum 2 numbers to 0 and two ways to 1, 3 ways to 2, and so on. At 9, there are 10 possible ways. Recall that only integers between 0-9 are valid. Now observe how there is 1 way to to sum to 18 in this fashion (9+9), 2 ways to sum to 17, and so forth again (to optionally prove that this pattern holds, apply stars and bars up to 9 and notice the symmetry).
The answer then is the number of ways to write each component of each pair. This is or, since it's symmetrical between sum of 4 and 5, . Use summation rules to finally get .
~BJHHar
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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