A number has exactly 3 prime factors, 125 factors of this are perfect squares and 27 factors of this number are perfect cubes. Overall how many factors does this number have?

Solution (By Examveda Team)

Let a,b,c be three prime no.
Since the no. of perfect square as a factor is 125
Which can only be if every prime no. has power in the multiple of 2 including zeros (because any no. to the power of zero is a perfect square.)
So power must be 0,2,4,6,8
Thereby 5*5*5=125
Since no. of cube as a factor is 27
Therefore power in the prime no. will be in the multiple of 3
So power are 0,3,6
Hence the no. N = a^8*b^8*c^8
Hence total no. of factor is (8+1)(8+1)(8+1) = 729.

Alternatively,

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