Difference between revisions of "2016 AIME II Problems/Problem 11"
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==Solution== | ==Solution== | ||
− | 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>142+125-17=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>142+125-17=250</math>, so the desired answer is <math>999-250=\boxed{749}</math>. |
Solution by Shaddoll and firebolt360 | Solution by Shaddoll and firebolt360 | ||
==Solution 2== | ==Solution 2== | ||
− | 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> <math>1 \pmod{7}</math> and also <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>. | + | 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 == |
Latest revision as of 06:19, 16 August 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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