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

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~Lcz
 
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==Solution 2 (Generalized and Comprehensive)==
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===Claim===
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<b>Solution in progress. A million thanks for not editing.</b>
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~MRENTHUSIASM
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===Proof 1===
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<b>Solution in progress. A million thanks for not editing.</b>
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~MRENTHUSIASM
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===Proof 2===
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<b>Solution in progress. A million thanks for not editing.</b>
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~MRENTHUSIASM
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===Solution===
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<b>Solution in progress. A million thanks for not editing.</b>
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~MRENTHUSIASM
  
 
==See also==
 
==See also==
 
{{AIME box|year=2021|n=II|num-b=8|num-a=10}}
 
{{AIME box|year=2021|n=II|num-b=8|num-a=10}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 02:48, 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

Solution in progress. A million thanks for not editing.

~MRENTHUSIASM

Proof 1

Solution in progress. A million thanks for not editing.

~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

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