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avatar+489 

What is the greatest common divisor of 210011 and 210121?

 Sep 22, 2018

Best Answer 

 #2
avatar+6251 
+4

The theorem being used here is that

 

gcd(an1, am1)=agcd(m,n)1

 

in this case n=1001, m=1012

 

1001=711131012=21123so clearly gcd(1001, 1012)=112111=2047 as guest correctly computed

 

The proof of this theorem can be found online

 Sep 22, 2018
 #1
avatar
+3

I think the GCD of ((2^1001) - 1), ((2^1012) - 1) =2^(1012 - 1001) - 1 =2^11 - 1 =2,047

 Sep 22, 2018
 #2
avatar+6251 
+4
Best Answer

The theorem being used here is that

 

gcd(an1, am1)=agcd(m,n)1

 

in this case n=1001, m=1012

 

1001=711131012=21123so clearly gcd(1001, 1012)=112111=2047 as guest correctly computed

 

The proof of this theorem can be found online

Rom Sep 22, 2018

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