Latest TCS Aptitude Question SOLUTION: a number has 3 prime factors, 125 factors are perfect square, 27 factors are perfect cubes,find total number of factors

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(#M40052237) TCS Numerical Ability question number has 3 prime factors Keep an EYE

a number has 3 prime factors, 125 factors are perfect square, 27 factors are perfect cubes,find total number of factors

Asked In TCS jayan nair (9 years ago)

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KRISHNA KUMAR  (9 years ago)

We know that the total factors of a number N = ap.bq.cr ....
Now the total factors which are perfect squares of a number N = ([p2]+1).([q2]+1).([r2]+1)....
where [x] is greatest intezer less than that of x.
Given ([p2]+1).([q2]+1).([r2]+1).... = 125
So [p2]+1 = 5; [q2]+1 = 5; [p2]+1 = 5
[p2] = 4 ⇒ p = 8 or 9, similarly q = 8 or 9, r = 8 or 9
Given that 27 factors of this number are perfect cubes
so ([p3]+1).([q3]+1).([r3]+1).... = 27
So [p3]+1 = 3 ⇒ = [p3] = 2
⇒ p = 6, 7, 8
By combining we know that p = q = r = 8
So the given number should be in the format = a8.b8.c8 ....
Number of factors of this number = (8+1).(8+1).(8+1) = 729

Is this solution Helpfull? Yes (7) | No (2)  

Rani Pata  (8 years ago)

We know that the total factors of a number N = a^p.b^q.c^r ....
Now the total factors which are perfect squares of a number N = ([p/2]+1).([q/2]+1).([r/2]+1)....
where [x] is greatest intezer less than that of x.
Given ([p/2]+1).([q/2]+1).([r/2]+1).... = 125
So [p/2]+1 = 5; [q/2]+1 = 5; [p/2]+1 = 5
[p/2] = 4 ⇒ p = 8 or 9, similarly q = 8 or 9, r = 8 or 9
Given that 27 factors of this number are perfect cubes
so ([p/3]+1).([q/3]+1).([r/3]+1).... = 27
So [p/3]+1 = 3 ⇒ = [p/3] = 2
⇒ p = 6, 7, 8
By combining we know that p = q = r = 8
So the given number should be in the format = a^8.b^8.c^8 ....
Number of factors of this number = (8+1).(8+1).(8+1) = 729

Is this solution Helpfull? Yes (2) | No   

ponkia jay  (9 years ago)

729 is the answer

Is this solution Helpfull? Yes | No (10)  

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