Math,Gcd

Revision en1, by sahasumit288, 2016-10-15 22:25:19

n is a number and n=(p1^a) * (p2^b).Here p1 and p2 are prime.

gcd(i,n)=$\sum\limits_{i=1}^{p1^a}{}gcd(i, p1a) * \sum\limits_{i=1}^{p2^b}{}$ gcd(i,p2^b) How can I prove this?

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  Rev. Lang. By When Δ Comment
en5 English sahasumit288 2016-10-15 22:33:31 3 Tiny change: 'b)\n\nHow can to prove ' -> 'b)\n\nHow to prove '
en4 English sahasumit288 2016-10-15 22:29:28 2 Tiny change: 'gcd(i,b)\nHow can ' -> 'gcd(i,b)\n\nHow can '
en3 English sahasumit288 2016-10-15 22:28:23 79
en2 English sahasumit288 2016-10-15 22:26:07 3
en1 English sahasumit288 2016-10-15 22:25:19 216 Initial revision (published)