Difference between revisions of "2017 AIME I Problems/Problem 7"
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We begin as in solution 1 to rewrite the sum as <math>\binom{6}{a} \binom{6}{b} \binom{6}{c}</math> over all <math>a,b,c</math> such that <math>a+b+c=6</math>. | We begin as in solution 1 to rewrite the sum as <math>\binom{6}{a} \binom{6}{b} \binom{6}{c}</math> over all <math>a,b,c</math> such that <math>a+b+c=6</math>. | ||
− | Consider the polynomial <math>P(x)=(\binom{6}{0} + \binom{6}{1}x + \binom{6}{2}x^2 + \binom{6}{3}x^3 + \cdot \cdot \cdot + \binom{6}{6}x^6)^3</math>. | + | Consider the polynomial <math>P(x)=\left(\binom{6}{0} + \binom{6}{1}x + \binom{6}{2}x^2 + \binom{6}{3}x^3 + \cdot \cdot \cdot + \binom{6}{6}x^6\right)^3</math>. |
We can see the sum we wish to compute is just the coefficient of the <math>x^6</math> term. However <math>P(x)=((1+x)^6)^3=(1+x)^{18}</math>. Therefore, the coefficient of the <math>x^6</math> term is just <math>\binom{18}{6} = 18564</math> so the answer is <math>\boxed{564}</math>. | We can see the sum we wish to compute is just the coefficient of the <math>x^6</math> term. However <math>P(x)=((1+x)^6)^3=(1+x)^{18}</math>. Therefore, the coefficient of the <math>x^6</math> term is just <math>\binom{18}{6} = 18564</math> so the answer is <math>\boxed{564}</math>. | ||
Revision as of 13:30, 3 January 2018
Problem 7
For nonnegative integers and with , let . Let denote the sum of all , where and are nonnegative integers with . Find the remainder when is divided by .
Solution 1
Let , and note that . The problem thus asks for the sum over all such that . Consider an array of 18 dots, with 3 columns of 6 dots each. The desired expression counts the total number of ways to select 6 dots by considering each column separately. However, this must be equal to . Therefore, the answer is .
-rocketscience
Solution 2
Treating as , this problem asks for But can be computed through the following combinatorial argument. Choosing elements from a set of size is the same as splitting the set into two sets of size and choosing elements from one, from the other where . The number of ways to perform such a procedure is simply . Therefore, the requested sum is As such, our answer is .
- Awsomness2000
Solution 3 (Major Bash)
Case 1: .
Subcase 1: Subcase 2: Subcase 3:
Case 2:
By just switching and in all of the above cases, we will get all of the cases such that is true. Therefore, this case is also
Case 3:
Solution 4
We begin as in solution 1 to rewrite the sum as over all such that . Consider the polynomial . We can see the sum we wish to compute is just the coefficient of the term. However . Therefore, the coefficient of the term is just so the answer is .
- mathymath
See Also
2017 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 6 |
Followed by Problem 8 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.