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Difference between revisions of "2021 AIME II Problems/Problem 9"

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(Changed "Solution" to "Solution 1", and added "Solution 2", outlining a much simpler approach.)
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==Solution 2==
 
==Solution 2==
  
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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>.
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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\ne 1</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>. Note that <math>2^{2m} \equiv 1 \pmod{p}</math>. From this, we can say that <math>2m</math> and <math>n</math> are both multiples of <math>d</math>, but <math>m</math> is not. Thus, we have <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>. It is not too hard to see that our steps are reversible, and therefore this condition is necessary and sufficient.
  
 
The rest of the solution now proceeds as in Solution 1.
 
The rest of the solution now proceeds as in Solution 1.

Revision as of 00:18, 4 February 2024

Problem

Find the number of ordered pairs $(m, n)$ such that $m$ and $n$ are positive integers in the set $\{1, 2, ..., 30\}$ and the greatest common divisor of $2^m + 1$ and $2^n - 1$ is not $1$.

Solution 1

This solution refers to the Remarks section.

By the Euclidean Algorithm, we have \[\gcd\left(2^m+1,2^m-1\right)=\gcd\left(2,2^m-1\right)=1.\] We are given that $\gcd\left(2^m+1,2^n-1\right)>1.$ Multiplying both sides by $\gcd\left(2^m-1,2^n-1\right)$ gives \begin{align*} \gcd\left(2^m+1,2^n-1\right)\cdot\gcd\left(2^m-1,2^n-1\right)&>\gcd\left(2^m-1,2^n-1\right) \\ \gcd\left(\left(2^m+1\right)\left(2^m-1\right),2^n-1\right)&>\gcd\left(2^m-1,2^n-1\right) \hspace{12mm} &&\text{by }\textbf{Claim 1} \\ \gcd\left(2^{2m}-1,2^n-1\right)&>\gcd\left(2^m-1,2^n-1\right) \\ 2^{\gcd(2m,n)}-1&>2^{\gcd(m,n)}-1 &&\text{by }\textbf{Claim 2} \\ \gcd(2m,n)&>\gcd(m,n), \end{align*} which implies that $n$ must have more factors of $2$ than $m$ does.

We construct the following table for the first $30$ positive integers: \[\begin{array}{c|c|c} && \\ [-2.5ex] \boldsymbol{\#}\textbf{ of Factors of }\boldsymbol{2} & \textbf{Numbers} & \textbf{Count} \\ \hline && \\ [-2.25ex] 0 & 1,3,5,7,9,11,13,15,17,19,21,23,25,27,29 & 15 \\ && \\ [-2.25ex] 1 & 2,6,10,14,18,22,26,30 & 8 \\ && \\ [-2.25ex] 2 & 4,12,20,28 & 4 \\ && \\ [-2.25ex] 3 & 8,24 & 2 \\ && \\ [-2.25ex] 4 & 16 & 1 \\ \end{array}\] To count the ordered pairs $(m,n),$ we perform casework on the number of factors of $2$ that $m$ has:

  1. If $m$ has $0$ factors of $2,$ then $m$ has $15$ options and $n$ has $8+4+2+1=15$ options. So, this case has $15\cdot15=225$ ordered pairs.

  2. If $m$ has $1$ factor of $2,$ then $m$ has $8$ options and $n$ has $4+2+1=7$ options. So, this case has $8\cdot7=56$ ordered pairs.

  3. If $m$ has $2$ factors of $2,$ then $m$ has $4$ options and $n$ has $2+1=3$ options. So, this case has $4\cdot3=12$ ordered pairs.

  4. If $m$ has $3$ factors of $2,$ then $m$ has $2$ options and $n$ has $1$ option. So, this case has $2\cdot1=2$ ordered pairs.

Together, the answer is $225+56+12+2=\boxed{295}.$

~Lcz ~MRENTHUSIASM

Solution 2

Consider any ordered pair $(m,n)$ such that $\gcd(2^m+1, 2^n-1) > 1$. There must exist some odd number $p\ne 1$ such that $2^m \equiv -1 \pmod{p}$ and $2^n \equiv 1 \pmod{p}$. Let $d$ be the order of $2$ modulo $p$. Note that $2^{2m} \equiv 1 \pmod{p}$. From this, we can say that $2m$ and $n$ are both multiples of $d$, but $m$ is not. Thus, we have $v_2(n) \ge v_2(d)$ and $v_2(m) + 1 = v_2(d)$. Substituting the latter equation into the inequality before gives $v_2(n) \ge v_2(m)+1$. Since $v_2(n)$ and $v_2(m)$ are integers, this implies $v_2(n)>v_2(m)$. It is not too hard to see that our steps are reversible, and therefore this condition is necessary and sufficient.

The rest of the solution now proceeds as in Solution 1.

~Sedro

Remarks

Claim 1 (GCD Property)

If $\boldsymbol{r,s,}$ and $\boldsymbol{t}$ are positive integers such that $\boldsymbol{\gcd(r,s)=1,}$ then $\boldsymbol{\gcd(r,t)\cdot\gcd(s,t)=\gcd(rs,t).}$

As $r$ and $s$ are relatively prime (have no prime divisors in common), this property is intuitive.

~MRENTHUSIASM

Claim 2 (Olympiad Number Theory Lemma)

If $\boldsymbol{u,a,}$ and $\boldsymbol{b}$ are positive integers such that $\boldsymbol{u\geq2,}$ then $\boldsymbol{\gcd\left(u^a-1,u^b-1\right)=u^{\gcd(a,b)}-1.}$

There are two proofs to this claim, as shown below.

~MRENTHUSIASM

Claim 2 Proof 1 (Euclidean Algorithm)

If $a=b,$ then $\gcd(a,b)=a=b,$ from which the claim is clearly true.

Otherwise, let $a>b$ without the loss of generality. For all integers $p$ and $q$ such that $p>q>0,$ the Euclidean Algorithm states that \[\gcd(p,q)=\gcd(p-q,q)=\cdots=\gcd(p\operatorname{mod}q,q).\] We apply this result repeatedly to reduce the larger number: \[\gcd\left(u^a-1,u^b-1\right)=\gcd\left(u^a-1-u^{a-b}\left(u^b-1\right),u^b-1\right)=\gcd\left(u^{a-b}-1,u^b-1\right).\] Continuing, we have \begin{align*} \gcd\left(u^a-1,u^b-1\right)&=\gcd\left(u^{a-b}-1,u^b-1\right) \\ & \ \vdots \\ &=\gcd\left(u^{\gcd(a,b)}-1,u^{\gcd(a,b)}-1\right) \\ &=u^{\gcd(a,b)}-1, \end{align*} from which the proof is complete.

~MRENTHUSIASM

Claim 2 Proof 2 (Bézout's Identity)

Let $d=\gcd\left(u^a-1,u^b-1\right).$ It follows that $u^a\equiv1\pmod{d}$ and $u^b\equiv1\pmod{d}.$

By Bézout's Identity, there exist integers $x$ and $y$ such that $ax+by=\gcd(a,b),$ so \[u^{\gcd(a,b)}=u^{ax+by}=(u^a)^x\cdot(u^b)^y\equiv1\pmod{d},\] from which $u^{\gcd(a,b)}-1\equiv0\pmod{d}.$ We know that $u^{\gcd(a,b)}-1\geq d.$

Next, we notice that \begin{align*} u^a-1&=\left(u^{\gcd(a,b)}-1\right)\left(u^{a-\gcd{(a,b)}}+u^{a-2\gcd{(a,b)}}+u^{a-3\gcd{(a,b)}}+\cdots+1\right), \\ u^b-1&=\left(u^{\gcd(a,b)}-1\right)\left(u^{b-\gcd{(a,b)}}+u^{b-2\gcd{(a,b)}}+u^{b-3\gcd{(a,b)}}+\cdots+1\right). \end{align*} Since $u^{\gcd(a,b)}-1$ is a common divisor of $u^a-1$ and $u^b-1,$ we conclude that $u^{\gcd(a,b)}-1=d,$ from which the proof is complete.

~MRENTHUSIASM

Video Solution

https://youtu.be/h3awf7yhGZ4

~MathProblemSolvingSkills.com


Video Solution by Interstigation

https://youtu.be/kasgsb0Rge4

~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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