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If a number N can be represented as Pa1×Pb2×Pc3 where P1,P2,P3 are prime factors of N then number of factors of N can be found as (a+1)×(b+1)×(c+1) |
| Answer» If a number N can be represented as Pa1×Pb2×Pc3 where P1,P2,P3 are prime factors of N then number of factors of N can be found as (a+1)×(b+1)×(c+1) | |