Difference between revisions of "2021 AIME II Problems/Problem 3"
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− | + | Since <math>3</math> is one of the numbers, a product with a <math>3</math> in it is automatically divisible by <math>3</math>, so WLOG <math>x_3=3</math>, we will multiply by <math>5</math> afterward since any of <math>x_1, x_2, ..., x_5</math> would be <math>3</math>, after some cancelation we see that now all we need to find is the number of ways that <math>x_5x_1(x_4+x_2)</math> is divisible by <math>3</math>, since <math>x_5x_1</math> is never divisible by <math>3</math>, now we just need to find the number of ways <math>x_4+x_2</math> is divisible by <math>3</math>, after some calculation you will see that there are <math>16</math> ways to choose <math>x_1, x_2, x_4,</math> and <math>x_5</math> in this way. So the desired answer is <math>16 \times 5=\boxed{080}</math>. | |
==See also== | ==See also== | ||
{{AIME box|year=2021|n=II|num-b=2|num-a=4}} | {{AIME box|year=2021|n=II|num-b=2|num-a=4}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 14:14, 22 March 2021
Problem
Find the number of permutations of numbers such that the sum of five products
Solution
Since is one of the numbers, a product with a in it is automatically divisible by , so WLOG , we will multiply by afterward since any of would be , after some cancelation we see that now all we need to find is the number of ways that is divisible by , since is never divisible by , now we just need to find the number of ways is divisible by , after some calculation you will see that there are ways to choose and in this way. So the desired answer is .
See also
2021 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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