Difference between revisions of "2018 AMC 12B Problems/Problem 22"
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− | Suppose that <math>P(x)=ax^3+bx^2+cx+d.</math> This problem is equivalent to counting the ordered quadruples <math>(a,b,c,d),</math> where all of <math>a,b,c,</math> and <math>d</math> are integers from <math>0</math> through <math>9 | + | Suppose that <math>P(x)=ax^3+bx^2+cx+d.</math> This problem is equivalent to counting the ordered quadruples <math>(a,b,c,d),</math> where all of <math>a,b,c,</math> and <math>d</math> are integers from <math>0</math> through <math>9</math> such that <cmath>P(-1)=-a+b-c+d=-9.</cmath> Let <math>a'=9-a</math> and <math>c'=9-c.</math> Note that both of <math>a'</math> and <math>c'</math> are integers from <math>0</math> through <math>9.</math> Moreover, the ordered quadruples <math>(a,b,c,d)</math> and the ordered quadruples <math>(a',b,c',d)</math> have one-to-one correspondence. |
We rewrite the given equation as <math>(9-a)+b+(9-c)+d=9,</math> or <cmath>a'+b+c'+d=9.</cmath> By the stars and bars argument, there are <math>\binom{9+4-1}{4-1}=\boxed{\textbf{(D) } 220}</math> ordered quadruples <math>(a',b,c',d).</math> | We rewrite the given equation as <math>(9-a)+b+(9-c)+d=9,</math> or <cmath>a'+b+c'+d=9.</cmath> By the stars and bars argument, there are <math>\binom{9+4-1}{4-1}=\boxed{\textbf{(D) } 220}</math> ordered quadruples <math>(a',b,c',d).</math> |
Revision as of 10:38, 26 October 2021
Problem
Consider polynomials of degree at most , each of whose coefficients is an element of . How many such polynomials satisfy ?
Solution 1
Suppose that This problem is equivalent to counting the ordered quadruples where all of and are integers from through such that Let and Note that both of and are integers from through Moreover, the ordered quadruples and the ordered quadruples have one-to-one correspondence.
We rewrite the given equation as or By the stars and bars argument, there are ordered quadruples
~pieater314159 ~MRENTHUSIASM
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.