Latest TCS Aptitude Question SOLUTION: The prime factorization of intezer N is A x A x B x C, where A, B and C are all distinct prine intezers. How many factors does N have?

• Placement Papers
• /
• TCS
• /
• HR Interview
• /
• Interview

TCS Question Papers: TCS Interview Experience(90) TCS Company Profile

(#M40011172) TCS HR Interview question Keep an EYE

The prime factorization of intezer N is A x A x B x C, where A, B and C are all distinct prine intezers. How many factors does N have?

Asked In TCS krati (11 years ago)

Unsolved Read Solution (6)

Is this Puzzle helpful?   (22)   (15) Submit Your Solution Interview

Read Solution (6) :

Sunil jaiswal  (11 years ago)

Since A,B,C are the prime no.hence we can apply the formula:
no of factors=(n+1)(m+1)(p+1)......
where n,m,p...are the powers of prime no..here we can write the given expression as A^2 x B^1 x C^1. hence no of factors are(2+1)(1+1)(1+1)=12(Ans)

Is this solution Helpfull? Yes (65) | No (2)  

arumoy  (11 years ago)

Here n is A*A*B*C=A^2*B*C.so the no of factors of n is
(2+1)*(1+1)*(1+1)=3*2*2=12(ans).

Is this solution Helpfull? Yes (34) | No (8)  

sainath  (11 years ago)

given N=A*A*B*C AND A,B,C ARE PRIME INTEGERS...
SO A,B,C HAVE NO FACTORS EXCEPT 1 AND THEM SELVES...
the other factors of N are formed by selecting product of at least 2 of A*A*B*C
that is product of 2 numbers formed (A*A,A*B,A*C,B*C)=4
product of 3 numbers formed (A*A*B,A*A*C,A*B*C)=3
OTHER KNOWN FACTORS ARE (1,A,B,C,N)=5
SO TOTAL FACTORS OF N ARE 12....

Is this solution Helpfull? Yes (11) | No (7)  

sainath  (11 years ago)

given N=A*A*B*C AND A,B,C ARE PRIME INTEGERS...
SO A,B,C HAVE NO FACTORS EXCEPT 1 AND THEM SELVES...
the other factors of N are formed by selecting product of at least 2 of A*A*B*C
that is product of 2 numbers formed (A*A,A*B,A*C)=3
product of 3 numbers formed (A*A*B,A*A*C,A*B*C)=3
OTHER KNOWN FACTORS ARE (1,A,B,C,N)=5
SO TOTAL FACTORS OF N ARE 11....

Is this solution Helpfull? Yes (3) | No (18)  

debopama sarkar  (11 years ago)

factors are A,A,B,C,AA,AB,AC,AB,AC,AAB,AAC,AABC,BC,ABC,AAB,AAC,ABC,...(since all are prime factors so the two A's are different) ans is 24

Is this solution Helpfull? Yes (1) | No (20)  

Tridib Samanta  (4 years ago)

If n=(p1)a∗(p2)b∗(p3)c∗......where p1,p2,p3 are primes then, the total number of factors of n is (a+1)∗(b+1)∗(c+1)∗........

so, here, a,b,c are primes, and their powers are 2,1,1 respectively.

so, according to the formula,
number of factors of n is (2+1)∗(1+1)∗(1+1) which turns out to be 3∗2∗2=12.

Is this solution Helpfull? Yes | No   

Please Login to Read Solution.