2021 AIME II Problems/Problem 9

Revision as of 02:55, 1 April 2021 by MRENTHUSIASM (talk | contribs) (Claim)

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 $\gcd\left(u^a-1,u^b-1\right)=u^{\gcd(a,b)}-1.$

~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 (ProblemsAnswer KeyResources)
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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