Question

A number n has 3 prime factors and 27 of the factors are perfect cubes. if 125 of the factors of n are perfect squares how many factors does n have

Answer 1

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Answer 2

Answer:

729

Step-by-step explanation:

We know any number can be expressed in terms of their powers of prime factors

As the given no has three prime factors

N=(A^p)(B^q)(C^r)

Total no of factors of this no = (p+1)(q+1)(r+1)

No of square factors =([p/2] +1)([q/2] +1)([r/2] +1)

No of cubic factors =([p/3]+1)([q/3]+1)([r/3]+1)

where [x]= Z

but it is given that the no. has 125 square factors i.e

([p/2] +1)([q/2] +1)([r/2] +1)=125

=> [p/2]=4; [q/2]=4; [r/2]=4

=> p=q=r=8,9.

Also given no. of cubic factors is 27 i.e.,

([p/3] +1)([q/3] +1)([r/3] +1)=27

[p/3]+1=3,[q/3]+1=3 nd [r/2]+1=3

=> [p/3]=2=[q/3]=[r/3]

possible values of p,q,r will be 6,7 or 8

From both the answers we can conclude p=q=r=8

Total no of factors =(8+1)(8+1)(8+1)=729