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back to boardIdea 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 Posted by phizaz 10 Jun 2011 21:40 I think its too strong hint! |
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