Difference between revisions of "2021 AIME II Problems/Problem 9"
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Find the number of ordered pairs <math>(m, n)</math> such that <math>m</math> and <math>n</math> are positive integers in the set <math>\{1, 2, ..., 30\}</math> and the greatest common divisor of <math>2^m + 1</math> and <math>2^n - 1</math> is not <math>1</math>. | Find the number of ordered pairs <math>(m, n)</math> such that <math>m</math> and <math>n</math> are positive integers in the set <math>\{1, 2, ..., 30\}</math> and the greatest common divisor of <math>2^m + 1</math> and <math>2^n - 1</math> is not <math>1</math>. | ||
| − | ==Solution== | + | ==Solution 1== |
| − | We | + | We make use of the (olympiad nt) lemma that <math>\gcd(2^a-1,2^b-1)=2^{\gcd(a,b)}-1</math>. |
| + | Noting <math>\gcd(2^m+1,2^m-1)=\gcd(2^m+1,2)=1</math>, we have (by difference of squares)<cmath>\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) </cmath><cmath>\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).</cmath> It is now easy to calculate the answer (with casework on <math>\nu_2(m)</math>) as <math>15 \cdot 15+8 \cdot 7+4 \cdot 3+2 \cdot 1=\boxed{295}</math>. | ||
==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 17:56, 22 March 2021
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 nt) 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 
.
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. 
