A number has exactly 3 prime factors, 125 factors of this number are perfect squares and 27 factors of this number are perfect cubes. overall how many factors does the number have?
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its wrong question you say only 3 prime factor
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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
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