Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m
(→‎Proof 1 (Euclidean Algorithm): Made the proof concise.)
Line 22: Line 22:
 
Otherwise, let <math>a>b</math> without the loss of generality. Note that for all integers <math>p>q>0,</math> the Euclidean Algorithm states that <cmath>\gcd(p,q)=\gcd(p-q,q)=\cdots=\gcd(q,p\text{ mod }q).</cmath> We apply this result repeatedly to reduce the larger number: <cmath>\gcd\left(u^a-1,u^b-1\right)=\gcd\left(u^b-1,u^a-1-u^{a-b}\left(u^b-1\right)\right)=\gcd\left(u^b-1,u^{a-b}-1\right).</cmath>
 
Otherwise, let <math>a>b</math> without the loss of generality. Note that for all integers <math>p>q>0,</math> the Euclidean Algorithm states that <cmath>\gcd(p,q)=\gcd(p-q,q)=\cdots=\gcd(q,p\text{ mod }q).</cmath> We apply this result repeatedly to reduce the larger number: <cmath>\gcd\left(u^a-1,u^b-1\right)=\gcd\left(u^b-1,u^a-1-u^{a-b}\left(u^b-1\right)\right)=\gcd\left(u^b-1,u^{a-b}-1\right).</cmath>
 
Continuing, we will get
 
Continuing, we will get
−
<cmath>\begin{align*}
+
<math></math>\begin{align*}
 
\gcd\left(u^a-1,u^b-1\right)&=\cdots \\
 
\gcd\left(u^a-1,u^b-1\right)&=\cdots \\
 
&=\gcd\left(u^b-1,u^{a-b}-1\right) \\
 
&=\gcd\left(u^b-1,u^{a-b}-1\right) \\
 
&=\cdots \\
 
&=\cdots \\
−
&=\gcd\left(u^d-1,u^d-1\right) \\
+
&=\gcd\left(u^{\gcd(a,b)}-1,u^{\gcd(a,b)}-1\right) \\
−
&=u^d-1
+
&=u^{\gcd(a,b)}-1,
−
\end{align*}</cmath>
+
\end{align*}
−
for some positive integer <math>d,</math> where <math>d</math> is a common divisor of <math>a</math> and <math>b.</math>
+
from which the proof is complete.
−
 
 
−
From this equation, it is clear that <math>d</math> must be a linear combination of <math>a</math> and <math>b.</math> That is, there exist integers <math>x</math> and <math>y</math> for which <math>ax+by=d.</math> The smallest positive integer in the form of <math>ax+by</math> is <math>\gcd(a,b).</math> Therefore, We have <math>d=\gcd(a,b),</math> and the proof is complete.
 
  
 
~MRENTHUSIASM
 
~MRENTHUSIASM

Revision as of 03:24, 1 April 2021

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

We make use of the (olympiad number theory) lemma that $\gcd(2^a-1,2^b-1)=2^{\gcd(a,b)}-1$.

Noting $\gcd(2^m+1,2^m-1)=\gcd(2^m+1,2)=1$, we have (by difference of squares)\[\gcd(2^m+1,2^n-1) \neq 1 \iff \gcd(2^{2m}-1,2^n-1) \neq \gcd(2^m-1,2^n-1)\]\[\iff 2^{\gcd(2m,n)}-1 \neq 2^{\gcd(m,n)}-1 \iff \gcd(2m,n) \neq \gcd(m,n) \iff \nu_2(m)<\nu_2(n).\] It is now easy to calculate the answer (with casework on $\nu_2(m)$) as $15 \cdot 15+8 \cdot 7+4 \cdot 3+2 \cdot 1=\boxed{295}$.

~Lcz

Solution 2 (Generalized and Comprehensive)

Claim

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

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

~MRENTHUSIASM

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. Note that for all integers $p>q>0,$ the Euclidean Algorithm states that \[\gcd(p,q)=\gcd(p-q,q)=\cdots=\gcd(q,p\text{ mod }q).\] We apply this result repeatedly to reduce the larger number: \[\gcd\left(u^a-1,u^b-1\right)=\gcd\left(u^b-1,u^a-1-u^{a-b}\left(u^b-1\right)\right)=\gcd\left(u^b-1,u^{a-b}-1\right).\] Continuing, we will get $$ (Error compiling LaTeX. Unknown error_msg)\begin{align*} \gcd\left(u^a-1,u^b-1\right)&=\cdots \\ &=\gcd\left(u^b-1,u^{a-b}-1\right) \\ &=\cdots \\ &=\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

Proof 2

Solution in progress. A million thanks for not editing.

~MRENTHUSIASM

Solution

Solution in progress. A million thanks for not editing.

~MRENTHUSIASM

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2021_AIME_II_Problems/Problem_9&oldid=150723"