Thursday, 1 December 2016

number theory - How to compute these GCD's?



Please suggest me how to compute the GCD of theese really big numbers :





  1. GCD of $2^{120547564397}-1$ and $2^{356946681940}-1$

  2. GCD of $2^n-1$ and $n!$ where $n=3^{19}$



Thanks to Bill Dubuque's answer I understood that the first problem could be solved by the property that $gcd(f(m), f(n)) = f(gcd(m,n))$ if $ f(n) \equiv f(n-m) (mod\ f(m)),\ \ \ f(0)\ =\ 0$.



Any hints for the second one?


Answer




HINT $\rm\ (2^a-1,\:2^b-1)\ =\ 2^{(a,b)}-1\ $ where $\rm\ (a,b) := gcd(a,b)\:.\ $ For proofs see here, or here - which has a polynomial analog.


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