Latest TCS Aptitude Question SOLUTION: A Number has exactly 3 prime factors,125 factors of this number are perfect square and 27 factors of this number are perfect cubes,total how many factors does this number have?

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A Number has exactly 3 prime factors,125 factors of this number are perfect square and 27 factors of this number are perfect cubes,total how many factors does this number have?

Asked In TCS anirudh anand (11 years ago)

Solved Prem Batra Read Solution (7)

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Read Solution (7) :

Prem Batra  (11 years ago)

i think the answer is 729, here goes my logic

N = p1 ^ (x) * p2 ^ (y) * p3 ^ (z)

where p1, p2, p3 are primes and their powers are x, y and z

if i have to form a perfect cube i must choose x, y and z to be multiples of 3 starting from 0, 3, 6, 9, 12, ....
if i need 27 perfect cubes then each of the powers can take the following values 0, 3, 6 (3 possible values for the 3 powers)

hence we would be looking at N = p1 ^ (6) * p2 ^ (6) * p3 ^ (6)
these are 27 combinations = 27 perfect cubes
since there are 3 choices for every power ie. x, y, z the total number of combinations such that i choose a multiple of 3 for x, y, z is 3 * 3 * 3 = 27

Similar logic for the perfect squares
x, y, z must be multiples of 2 starting from 0, 2, 4, 6, 8, 1,....
if i need 125 perfect squares the each of the powers can take the following values 0, 2, 4, 6, 8 (5 posible values for the 3 powers)
hence we would be looking at N = p1 ^ ( * p2 ^ ( * p3 ^ (
thse are 125 combinations = 125 perfet squares

since this is the larger case which includes the case for perfect cubes i'll asume
N = p1 ^ ( * p2 ^ ( * p3 ^ (

total number of factors = (8 + 1) * (8 + 1) * (8 + 1) = 9 * 9 * 9 = 729

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Vinay Sipani   (11 years ago)

After a series of calculations,I got the following shortcut !!!

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

3b = (2a-1) or 2a or (2a+2)

Number of factors of the number = (2a)^n or (2a - 1)^n

Here a = 5 , b = 3 , n = 3
=> number of factors = either 9^3 or 10^3
= 729 or 1000

Is this solution Helpfull? Yes (4) | No (3)  

Vinay Sipani   (11 years ago)

number of factors = either (2a)^n or (2a-1)^n 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 2(a-1) > 3(b-1)
=> no of factors = (2a-1)^n = 9^3= 729

Is this solution Helpfull? Yes (4) | No (2)  

raja  (11 years ago)

ANS IS 729

Is this solution Helpfull? Yes (3) | No (2)  

Sourav Dutta  (10 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-10^3=[(2*5)-1]^3=9^3=729.

Is this solution Helpfull? Yes (3) | No   

Shiva  (10 years ago)

Thanks Vinay Sipani...d short cut works nicely.....

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

Alok Jain  (11 years ago)

is dere any short-cut method 4 dis???

Is this solution Helpfull? Yes | No (2)  

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