The prime factorization of integer N is A*A*B*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

• n is A*A*B*C =A^2*B*C
  no. of factors=(2+1)(1+1)(1+1)
  =12
  REASON:- A HAS POWER OF 2,B HAS POWER OF 1,C HAS POWER 1
  SO PRIME FACTORIZATION CAN BE CALCULATED AS:
  IF A^P+B^Q+C^R,THEN
  PRIME FACTORIZATION IS (P+1)*(Q+1)*(R+1)
  SO IN THIS CASE P=2,Q=1,R=1
  SO (2+1)(1+1)(1+1)=12 FACTORS(ANS)
• 11 years agoHelpfull: Yes(83) No(2)
• n is A*A*B*C =A^2*B*C
  no. of factors=(2+1)(1+1)(1+1)
  =12
• 11 years agoHelpfull: Yes(7) No(5)
• 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(3) No(2)
• IN THIS QUES WE CAL ALSO TAKE
  A=2,B=1,C=3
  THEN 2^2*1*3=12 FACTORS
• 11 years agoHelpfull: Yes(2) No(5)
• 12
  
  N will have 3*2*2 = 12 factors.
• 11 years agoHelpfull: Yes(1) No(9)
• 12 factors.
  3+3+3+3=12 factors.
• 11 years agoHelpfull: Yes(1) No(4)
• can you please explain the solution clearly..@parag kamra
• 11 years agoHelpfull: Yes(1) No(3)
• A=1,B=2,C=3
  (1+1)*2*3=12 FACTORS
• 11 years agoHelpfull: Yes(1) No(8)
• thank you..@parag kamra
• 11 years agoHelpfull: Yes(1) No(8)
• if n=a^p*b^q*c^r
  then no. of factors=(p+1)(q+1)(r+1)-1
  so (2+1)(1+1)(1+1)-1= 11 ans
• 11 years agoHelpfull: Yes(0) No(4)
• GIVEN N=A*A*B*C
  N=(A^2)*(B^1)*(C^1)
  
  PRIME FACTORIZATION OF N= (A^P)*(B^Q)*(C^R) WHERE P,Q,R ARE PRIME FACTORS
  NUMBER OF FACTORS OF N WILL BE EXPRESSED AS = (P+1)*(Q+1)*(R+1)
  HERE P=2, Q=1, R=1
  (2+1)*(1+1)*(1+1)=12
  ANS-: 12
• 8 years agoHelpfull: Yes(0) No(1)
• answer 24
  
  2*(!4/!2)
  
• 8 years agoHelpfull: Yes(0) No(1)