Latest TCS Aptitude Question SOLUTION: A number has exactly 3 prime factors, 125 factors of this no. Are perfect square and 27 factors of this no. Are perfect cubes. Overall how many factors does the number have ?

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A number has exactly 3 prime factors, 125 factors of this no. Are perfect square and 27 factors of this no. Are perfect cubes. Overall how many factors does the number have ?

Asked In TCS shubham (9 years ago)

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priyanka  (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 (10) | No (2)  

Sourick  (9 years ago)

(3^125)x(2^57)

Is this solution Helpfull? Yes (5) | No (1)  

anitha thummalapally  (9 years ago)

If a number has n prime factors , number of factors that are perfect squares = a^n and number of factors that are perfect cubes = b^n then
number of factors = based on following conditions

If 2(a-1) > 3(b-1) , then (2a-1)^n
If 2(a-1) < 3(b-1) , then (2a)^n
If 2(a-1) = 3(b-1) , then both are possible.
here no of prime factors=n=3
so, a^3=125 ; a=5 ; 2(a-1)=8 and
b^3=27 ; b=3 ;3(b-1)=6
therefore 2(a-1)> 3(b-1)
so as per the condition mention above
the total no of factors=(2a-1)^3=[(2*5)-1]^3=9^3=729.

Is this solution Helpfull? Yes (5) | No   

prerna singh  (9 years ago)

all answers are different .Please provide a correct solution.

Is this solution Helpfull? Yes (4) | No   

Abhishek nkumar  (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 (4) | No   

santosh bagathi  (9 years ago)

334

prime factors=3
squares(x^2=x*x)=125=> 125+125=250
cubes(x^3=x*x*x)=27 => 27+27+27=81

total factors = 3+250+81 = 334

Is this solution Helpfull? Yes | No (12)  

ronnie singh  (9 years ago)

a number has aform like a^p *b^q*c^r and so on den the total no of factors for this would be (p+1)(q+1)(r+1) nw in dis question no has 125 perfect sqr that is 125 a^p so total factor from dis is 125*(p+1) here p=2 n coms out to b 125* 3=375 an for cube v have 27*4=108 total will b 483

Is this solution Helpfull? Yes | No (3)  

Rishabh Kaushik  (9 years ago)

let the number is n
now n=a*b*c* a1^2*a2^2........*a125^2 *b1^3*b2^3........b27^3
where a,b,c are for three prime numbers.
a1,a2,a3.......a125 for perfect squares.
b1,b2,b3........b27 for perfect cubes.
now so thats why total number of factors are 2^3* 3^125* 4^27

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