Difference between revisions of "2021 AIME II Problems/Problem 9"
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Find the number of ordered pairs <math>(m, n)</math> such that <math>m</math> and <math>n</math> are positive integers in the set <math>\{1, 2, ..., 30\}</math> and the greatest common divisor of <math>2^m + 1</math> and <math>2^n - 1</math> is not <math>1</math>. | Find the number of ordered pairs <math>(m, n)</math> such that <math>m</math> and <math>n</math> are positive integers in the set <math>\{1, 2, ..., 30\}</math> and the greatest common divisor of <math>2^m + 1</math> and <math>2^n - 1</math> is not <math>1</math>. | ||
− | ==Solution== | + | ==Solution 1== |
This solution refers to the <b>Remarks</b> section. | This solution refers to the <b>Remarks</b> section. | ||
Line 42: | Line 42: | ||
~Lcz ~MRENTHUSIASM | ~Lcz ~MRENTHUSIASM | ||
+ | |||
+ | ==Solution 2== | ||
+ | |||
+ | Consider any ordered pair <math>(m,n)</math> such that <math>\gcd(2^m+1, 2^n-1)>1</math>. There must exist some odd number <math>p</math> such that <math>2^m \equiv -1 \pmod{p}</math> and <math>2^n \equiv 1 \pmod{p}</math>. Let <math>d</math> be the order of <math>2</math> modulo <math>p</math>. Also note that <math>2^{2m} \equiv 1 \pmod{p}</math>. Thus, <math>2m</math> and <math>n</math> are both multiples of <math>d</math>, but <math>m</math> is not. Thus, we can say that <math>v_2(n) \ge v_2(d)</math> and <math>v_2(m) + 1 = v_2(d)</math>. Substituting the latter equation into the inequality before gives <math>v_2(n) \ge v_2(m)+1</math>. Since <math>v_2(n)</math> and <math>v_2(m)</math> are integers, this implies <math>v_2(n)>v_2(m)</math>. | ||
+ | |||
+ | The rest of the solution now proceeds as in Solution 1. | ||
+ | |||
+ | ~Sedro | ||
+ | |||
==Remarks== | ==Remarks== | ||
+ | |||
===Claim 1 (GCD Property)=== | ===Claim 1 (GCD Property)=== | ||
<b>If <math>\boldsymbol{r,s,}</math> and <math>\boldsymbol{t}</math> are positive integers such that <math>\boldsymbol{\gcd(r,s)=1,}</math> then <math>\boldsymbol{\gcd(r,t)\cdot\gcd(s,t)=\gcd(rs,t).}</math></b> | <b>If <math>\boldsymbol{r,s,}</math> and <math>\boldsymbol{t}</math> are positive integers such that <math>\boldsymbol{\gcd(r,s)=1,}</math> then <math>\boldsymbol{\gcd(r,t)\cdot\gcd(s,t)=\gcd(rs,t).}</math></b> |
Revision as of 23:58, 3 February 2024
Contents
Problem
Find the number of ordered pairs such that
and
are positive integers in the set
and the greatest common divisor of
and
is not
.
Solution 1
This solution refers to the Remarks section.
By the Euclidean Algorithm, we have
We are given that
Multiplying both sides by
gives
which implies that
must have more factors of
than
does.
We construct the following table for the first positive integers:
To count the ordered pairs
we perform casework on the number of factors of
that
has:
- If
has
factors of
then
has
options and
has
options. So, this case has
ordered pairs.
- If
has
factor of
then
has
options and
has
options. So, this case has
ordered pairs.
- If
has
factors of
then
has
options and
has
options. So, this case has
ordered pairs.
- If
has
factors of
then
has
options and
has
option. So, this case has
ordered pairs.
Together, the answer is
~Lcz ~MRENTHUSIASM
Solution 2
Consider any ordered pair such that
. There must exist some odd number
such that
and
. Let
be the order of
modulo
. Also note that
. Thus,
and
are both multiples of
, but
is not. Thus, we can say that
and
. Substituting the latter equation into the inequality before gives
. Since
and
are integers, this implies
.
The rest of the solution now proceeds as in Solution 1.
~Sedro
Remarks
Claim 1 (GCD Property)
If and
are positive integers such that
then
As and
are relatively prime (have no prime divisors in common), this property is intuitive.
~MRENTHUSIASM
Claim 2 (Olympiad Number Theory Lemma)
If and
are positive integers such that
then
There are two proofs to this claim, as shown below.
~MRENTHUSIASM
Claim 2 Proof 1 (Euclidean Algorithm)
If then
from which the claim is clearly true.
Otherwise, let without the loss of generality. For all integers
and
such that
the Euclidean Algorithm states that
We apply this result repeatedly to reduce the larger number:
Continuing, we have
from which the proof is complete.
~MRENTHUSIASM
Claim 2 Proof 2 (Bézout's Identity)
Let It follows that
and
By Bézout's Identity, there exist integers and
such that
so
from which
We know that
Next, we notice that
Since
is a common divisor of
and
we conclude that
from which the proof is complete.
~MRENTHUSIASM
Video Solution
~MathProblemSolvingSkills.com
Video Solution by Interstigation
~Interstigation
See Also
2021 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 8 |
Followed by Problem 10 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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