Difference between revisions of "2016 AIME II Problems/Problem 11"
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− | All integers <math>a</math> will have factorization <math>2^a3^b5^c7^d...</math>. Therefore, the number of factors in <math>a^7</math> is <math>(7a+1)(7b+1)...</math>, and for <math>a^8</math> is <math>(8a+1)(8b+1)...</math>. The most salient step afterwards is to realize that all numbers <math>N</math> | + | All integers <math>a</math> will have factorization <math>2^a3^b5^c7^d...</math>. Therefore, the number of factors in <math>a^7</math> is <math>(7a+1)(7b+1)...</math>, and for <math>a^8</math> is <math>(8a+1)(8b+1)...</math>. The most salient step afterwards is to realize that all numbers <math>N</math> not <math>1 \pmod{7}</math> and also not <math>1 \pmod{8}</math> satisfy the criterion. The cycle repeats every <math>56</math> integers, and by PIE, <math>7+8-1=14</math> of them are either <math>7</math>-nice or <math>8</math>-nice or both. Therefore, we can take <math>\frac{42}{56} * 1008 = 756</math> numbers minus the <math>7</math> that work between <math>1000-1008</math> inclusive, to get <math>\boxed{749}</math> positive integers less than <math>1000</math> that are not nice for <math>k=7, 8</math>. |
== See also == | == See also == |
Revision as of 17:32, 8 July 2024
Contents
Problem
For positive integers and , define to be -nice if there exists a positive integer such that has exactly positive divisors. Find the number of positive integers less than that are neither -nice nor -nice.
Solution
We claim that an integer is only -nice if and only if . By the number of divisors formula, the number of divisors of is . Since all the s are divisible by in a perfect power, the only if part of the claim follows. To show that all numbers are -nice, write . Note that has the desired number of factors and is a perfect kth power. By PIE, the number of positive integers less than that are either or is , so the desired answer is .
Solution by Shaddoll and firebolt360
Solution 2
All integers will have factorization . Therefore, the number of factors in is , and for is . The most salient step afterwards is to realize that all numbers not and also not satisfy the criterion. The cycle repeats every integers, and by PIE, of them are either -nice or -nice or both. Therefore, we can take numbers minus the that work between inclusive, to get positive integers less than that are not nice for .
See also
2016 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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