Difference between revisions of "2021 AIME II Problems/Problem 9"
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===Claim=== | ===Claim=== | ||
If <math>u,a,</math> and <math>b</math> are positive integers for which <math>u\geq2,</math> then <math>\gcd\left(u^a-1,u^b-1\right)=u^{\gcd(a,b)}-1.</math> | If <math>u,a,</math> and <math>b</math> are positive integers for which <math>u\geq2,</math> then <math>\gcd\left(u^a-1,u^b-1\right)=u^{\gcd(a,b)}-1.</math> | ||
| + | |||
| + | There are two proofs to this claim, as shown below. | ||
~MRENTHUSIASM | ~MRENTHUSIASM | ||
Revision as of 02:32, 1 April 2021
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
We make use of the (olympiad number theory) lemma that 
.
Noting 
, we have (by difference of squares)![\[\gcd(2^m+1,2^n-1) \neq 1 \iff \gcd(2^{2m}-1,2^n-1) \neq \gcd(2^m-1,2^n-1)\]](http://latex.artofproblemsolving.com/8/6/b/86b625ac2e3cbb47a97de575fb28b6003c304adf.png)
![\[\iff 2^{\gcd(2m,n)}-1 \neq 2^{\gcd(m,n)}-1 \iff \gcd(2m,n) \neq \gcd(m,n) \iff \nu_2(m)<\nu_2(n).\]](http://latex.artofproblemsolving.com/6/9/7/6978a078c3b5eb568b1696ff4832647563c8fcdf.png)
It is now easy to calculate the answer (with casework on 
) as 
.
~Lcz
Solution 2 (Generalized and Comprehensive)
Claim
If 
and 
are positive integers for which 
then 
There are two proofs to this claim, as shown below.
~MRENTHUSIASM
Proof 1 (Euclidean Algorithm)
If 
then 
from which the claim is clearly true.
Otherwise, let 
without the loss of generality. Note that for all integers 
we have ![\[\gcd(p,q)=\gcd(p-q,q).\]](http://latex.artofproblemsolving.com/9/5/d/95dd9f7d826b4a1e2fc5127272030d6fd197b8b6.png)
We apply this result repeatedly to reduce the larger number: ![\[\gcd\left(u^a-1,u^b-1\right)=\gcd\left(u^b-1,u^a-1-u^{a-b}\left(u^b-1\right)\right)=\gcd\left(u^b-1,u^{a-b}-1\right).\]](http://latex.artofproblemsolving.com/b/e/a/bea17e9902d91a311d63c167d2b71b42c7140059.png)
~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. 
