Prime factorization of integer n is A*A*B*C where A,B & C are called distinct prime integers.How many factors does N have?

• N=A*A*B*C =A^2*B*C
  
  NO. of factors of N = (2+1)*(1+1)*(1+1)=3*2*2 =12
  
• 10 years agoHelpfull: Yes(26) No(1)
• a^2*b^1*c^1 is the factors of number N:
  then number of factor= (2+1)*(1+1)*(1+1)
  = 12
• 10 years agoHelpfull: Yes(1) No(0)
• can u plz explain the solution clearly
  
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• (2+1)*(1+1)*(1+1)
  3*2*2=12
  
• 10 years agoHelpfull: Yes(0) No(0)