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?

• if N=ap+br+cr
  then prime factor of N will be: (p+1)(q+1)(r+!)
  here
  p=2,q=1,r=1
  (2+1)(1+1)(1+1)
  3*2*2
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• 3*2*2=12 total 12.
  formula number of factors =(p+1)(q+1)(c+1).... where p,q,r.. are A^p, B^q.C^r.....
  
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• A^2BC
  No. of factors= (2+1)*(1+1)*(1+1)=12
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• 3is correct
  
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• N=A^2*B*C so, no. of factors will be= (2+1)*(1+1)*(1+1)= 3*2*2=12
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• no.of factors of a number N=a^p*b^q*c^r...........=(p+1)*(q+1)*(r+1)............
  by applying this formula to the given question:
  N=A^2*B*C=(2+1)(1+1)(1+1)=12
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• N=A^2*B^1*C^1
  factors=(1+2)*(1+1)*(1+1)=3*2*2=12
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• 3*2*2 by formula (p+1)(q+1)(r+1)
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• 12 factor
  
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• A^2xbxc=(2+1)x(2)x(2)
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• (2+1)*(1+1)*(1+1)=12
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• prime factor of N will be: (p+1)(q+1)(r+1)
  then
  ans=3*2*2
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• no of factors of N=3*2*2=12
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• here simple formula may use to get prime factorisation = (p+1)(q+1)(r+1)
  hence, (2+1)(1+1)(1+1)=12
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• A^2 * B *C
  No.of prime factrs=(power+1)*(power+1)*......
  So,(2+1)*(1+1)*(1+1)
  =3*2*2
  =12
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