The prime factorization of 2.intezer N is A x A x B x C, where A, B and C are all distinct prine intezers. How many factors does N have?

• Here n is A*A*B*C=A^2*B*C.so the no of factors of n is (2+1)*(1+1)*(1+1)=3*2*2=12(ans).
• 11 years agoHelpfull: Yes(44) No(6)