the prime factorization of integer N is A*A*B*C,WHERE A,B,C are prime nos.How many factors does N have??

• the prime factorization of integer n is (a^p)*(b^q)*(c^r)
  
  so the number of factors of n will be expressed by the formula (p+1)*(q+1)*(r+1)
  
  given the prime factorization of integer n is (a^2)*(b^1)*(c^1)
  so the number of factors of n = (2+1)*(1+1)*(1+1) = 3*2*2 = 12
  
• 11 years agoHelpfull: Yes(21) No(0)
• A=1,B=2,C=3
  (1+1)*2*3=12 FACTORS(ANS)
• 11 years agoHelpfull: Yes(18) No(3)
• N=A^2*B*C
  no. of factors=(2+1)(1+1)(1+1)
  =12
• 11 years agoHelpfull: Yes(9) No(2)
• N will have 3*2*2 = 12 factors.
• 11 years agoHelpfull: Yes(2) No(7)
• nt able to understand :(
• 11 years agoHelpfull: Yes(1) No(0)
• plz explain
• 11 years agoHelpfull: Yes(0) No(1)
• =4!/2!
  permutation with 2 no. alike
  
• 11 years agoHelpfull: Yes(0) No(4)