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> not <math>1 | + | 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 == | ||
{{AIME box|year=2016|n=II|num-b=10|num-a=12}} | {{AIME box|year=2016|n=II|num-b=10|num-a=12}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 11:57, 26 July 2019
Contents
[hide]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
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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