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Number System
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
Read Solution (Total 12)
-
- 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) - 12 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 - 12 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 - 12 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 - 12 years agoHelpfull: Yes(2) No(5)
- 12
N will have 3*2*2 = 12 factors. - 12 years agoHelpfull: Yes(1) No(9)
- 12 factors.
3+3+3+3=12 factors. - 12 years agoHelpfull: Yes(1) No(4)
- can you please explain the solution clearly..@parag kamra
- 12 years agoHelpfull: Yes(1) No(3)
- A=1,B=2,C=3
(1+1)*2*3=12 FACTORS - 12 years agoHelpfull: Yes(1) No(8)
- thank you..@parag kamra
- 12 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 - 9 years agoHelpfull: Yes(0) No(1)
- answer 24
2*(!4/!2)
- 9 years agoHelpfull: Yes(0) No(1)
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