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) 12
b) 24
c) 4
d) 6

• as there r 2 A's (2+1)
  for 1 B (1+1)
  for 1 C (1+1)
  multiply them (2+1)*(1+1)*(1+1)=12
• 12 years agoHelpfull: Yes(23) No(6)
• by direct formula:-
  total no of factors of the A^p*B^q*C^r is (p+1)*(q+1)*(r+1)..
  so the answer is N^2*B*C = (2+1)*(1+1)*(1+1)= 3*2*2=12....so the ans is 12
• 12 years agoHelpfull: Yes(7) No(1)
• since the no has only prime factors. . it can not be divided by the any other integer expect them. . so factors of N are A,B,C & 1 :)
  THUS the ans will be 4 :)
• 12 years agoHelpfull: Yes(3) No(8)
• ans is 12 as A^2*B*C has factor (2+1)(1+1)(1+)=12
• 12 years agoHelpfull: Yes(2) No(0)
• LETS DO IT BY TAKING SIMPLE EXAMPLE
  Lets A,B,C to be simple prime factors as A=2,B=3,C=5
  A X A X B X C=60
  FACTORS OF 60=1,2,3,4,5,6,10.12,15,20,30,60 i.e. TOTAL=12
• 10 years agoHelpfull: Yes(2) No(0)
• 12 is correct...............
• 12 years agoHelpfull: Yes(0) No(0)
• y (2+1),(1+1),(1+1)
• 10 years agoHelpfull: Yes(0) No(0)