|
|
вернуться в форумIdea I've just figured out that if a==p1^k1*p2^k2*...*pn^kn(where p_i are prime numbers) then (k1+1)*(k2+1)*...*(kn+1) amount of numbers devide a; so you can use this idea in this problem. Re: Idea Послано phizaz 10 июн 2011 21:40 I think its too strong hint! |
|
|