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