The prime factorization of integer N is A x A x B x C, where A, B and C are all distinct prime integers. How many factors does N have?

• A*A*B*C=A^2*B*C
  Number of factor=(2+1)*(1+1)*(1+1)=12
• 8 years agoHelpfull: Yes(5) No(1)
• N= A*A*B*C or A^2*B*C
  no. of factors= (2+1)*(1+1)*(1+1)=12
• 8 years agoHelpfull: Yes(2) No(2)
• Using formula n=p1*p2....*pn
  Where n is number and p1,p2 are distinct prime no
  No of factor = (a1+1)*(a2+1).....*(an+1)
  where a is the power of respective prime number
  So the no is 3*2*2=12
• 8 years agoHelpfull: Yes(2) No(1)