The prine factorization of integers N is A*A*B*C,Where A,B,C are all distinct prime integers .how many factors does N have?

• there Is one formula for calculating total num. of factors.
  which is (p+1)*(q+1)*(r+1)......., where p , q , r are the number of time the distinct prime factors repeat.
  so, (2+1)(1+1) (1+1) = 12.
• 11 years agoHelpfull: Yes(13) No(0)
• ans :12
  put a=2,b=3,c=5 so Number will be 60;and it has 12 factors...u can choose any prime values for a,b,c answer will be same
• 11 years agoHelpfull: Yes(5) No(1)
• N=A*A*B*C=>A^2*B*C
  let it be of d form => N=(A^p)*(B^q)*(C^r); A,B,C are prime
  then the number of factors can be given as => (p+1)*(q+1)*(r+1)
  so no of factors will be => (2+1)*(1+1)*(1+1) => 12
• 11 years agoHelpfull: Yes(3) No(0)
• Total no of factors=(1+1)(1+1)(1+1)(1+1)=16
• 11 years agoHelpfull: Yes(1) No(6)
• if A,B,C are prime number then
  total factor of N is = (2+1)*(1+1)*(1+1)=12
• 11 years agoHelpfull: Yes(1) No(0)
• Number of prime factors of N is sum of powers of each factor
  i.e (since A comes twice the power of A becomes 2)=2+1+1=4
• 11 years agoHelpfull: Yes(0) No(1)
• (A+1)*(B+1)*(C+1) will give us the no. of factors..
  Here A appears twice..So (2+1)*(1+1)*(1+1)=12
• 11 years agoHelpfull: Yes(0) No(0)
• 12 factors.. 3*2*2=12
• 11 years agoHelpfull: Yes(0) No(0)