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

• N = A*A*B*C*C = A^2*B^1*C^2
  
  total no. of factors of N = (2+1)*(1+1)*(2+1)= 3*2*3 = 18
• 10 years agoHelpfull: Yes(42) No(0)
• 5
  2+2+1=5
  
• 10 years agoHelpfull: Yes(0) No(3)
• only 3 factors a,b,c
• 10 years agoHelpfull: Yes(0) No(3)