Difference between revisions of "2016 AIME II Problems/Problem 11"
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We claim that an integer <math>N</math> is only <math>k</math>-nice if and only if <math>N \equiv 1 \pmod k</math>. By the number of divisors formula, the number of divisors of <math>\prod_{i=1}^n p_i^{a_i}</math> is <math>\prod_{i=1}^n (a_i+1)</math>. Since all the <math>a_i</math>s are divisible by <math>k</math> in a perfect <math>k</math> power, the only if part of the claim follows. To show that all numbers <math>N \equiv 1 \pmod k</math> are <math>k</math>-nice, write <math>N=bk+1</math>. Note that <math>2^{kb}</math> has the desired number of factors and is a perfect kth power. By PIE, the number of positive integers less than <math>1000</math> that are either <math>1 \pmod 7</math> or <math>1\pmod 8</math> is <math>143+125-18=250</math>, so the desired answer is <math>999-250=\boxed{749}</math>. | We claim that an integer <math>N</math> is only <math>k</math>-nice if and only if <math>N \equiv 1 \pmod k</math>. By the number of divisors formula, the number of divisors of <math>\prod_{i=1}^n p_i^{a_i}</math> is <math>\prod_{i=1}^n (a_i+1)</math>. Since all the <math>a_i</math>s are divisible by <math>k</math> in a perfect <math>k</math> power, the only if part of the claim follows. To show that all numbers <math>N \equiv 1 \pmod k</math> are <math>k</math>-nice, write <math>N=bk+1</math>. Note that <math>2^{kb}</math> has the desired number of factors and is a perfect kth power. By PIE, the number of positive integers less than <math>1000</math> that are either <math>1 \pmod 7</math> or <math>1\pmod 8</math> is <math>143+125-18=250</math>, so the desired answer is <math>999-250=\boxed{749}</math>. | ||
− | Solution by Shaddoll | + | Solution by Shaddoll and firebolt360 |
==Solution 2== | ==Solution 2== |
Latest revision as of 13:27, 31 May 2020
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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