Difference between revisions of "2021 AIME II Problems/Problem 9"
MRENTHUSIASM (talk | contribs) (→Proof 1 (Euclidean Algorithm): Made the proof concise.) |
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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*} |
\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) \\ | ||
| Line 28: | Line 28: | ||
&=\gcd\left(u^{\gcd(a,b)}-1,u^{\gcd(a,b)}-1\right) \\ | &=\gcd\left(u^{\gcd(a,b)}-1,u^{\gcd(a,b)}-1\right) \\ | ||
&=u^{\gcd(a,b)}-1, | &=u^{\gcd(a,b)}-1, | ||
| − | \end{align*} | + | \end{align*}</cmath> |
from which the proof is complete. | from which the proof is complete. | ||
Revision as of 03:24, 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 
the Euclidean Algorithm states that ![\[\gcd(p,q)=\gcd(p-q,q)=\cdots=\gcd(q,p\text{ mod }q).\]](http://latex.artofproblemsolving.com/1/7/3/173d038677a8d7e0d68c4ee12ce6ae9f7092abca.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)
Continuing, we will get
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. 
