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== | ||
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
