The prime factorization of a number N is AxAxBxC, where A,B,C are all distinct prime integers. How many factors does N have?

• A has power of 2,B has power of 1,C has power 1
  prime factorization can be calculated
  if A^P+B^Q+C^R,den
  prime factorization z (P+1)*(Q+1)*(R+1)
  in dis ques P=2,Q=1,R=1
  So (2+1)(1+1)(1+1)=12
• 12 years agoHelpfull: Yes(20) No(1)
• given that n is A*A*B*C =A^2*B*C
  no. of factors=(2+1)(1+1)(1+1)
  =12
• 12 years agoHelpfull: Yes(8) No(0)
• sorry for a number with prime factorization AxAxBxBxC it will be (2+1)*(2+1)*(1+1)=18 right???
• 12 years agoHelpfull: Yes(2) No(3)
• plz explain hw did u come to this formula??? and suppose i have a number whose prime factorization is AxAxBxC then how many factors will it have???
• 12 years agoHelpfull: Yes(0) No(4)
• sorry i meant could u plz explain how u got this formula??? and if i have a number whose prime factorization is AxAxBxBxC then how many factors will it have???
• 12 years agoHelpfull: Yes(0) No(3)
• So that means for a number with prime factorization of AxAxBxBxC the factors will be=(2+1)*(2+1)*(1+1)*(1+1)=3*3*2*2=36????
• 12 years agoHelpfull: Yes(0) No(3)
• 12
  (2+1)(1+1)(1+1)
• 12 years agoHelpfull: Yes(0) No(1)
• 2+1)(1+1)(1+1)=12
• 12 years agoHelpfull: Yes(0) No(1)