Question

What is the smallest positive integer $n$ such that $55n^{3}$ has exactly 55 positive divisors (including 1 and itself)?

Answer 1

We know That total No. of factors  

=product of (prime no  

′

s power+1)

If N is the number of different divisors:

N=(p  

1

​  

+1)⋅(p  

2

​  

+1)⋅⋅⋅(p  

n

​  

+1)

100=2  

2

×5  

2

 

=2×2×5×5=(1+1)(1+1)(4+1)(4+1)

Then the integer n=  a  

1

p  

1

​  

 

​  

⋅a  

2

p  

2

​  

 

​  

⋅⋅⋅⋅a  

n

p  

n

​  

 

​  

 

For the smallest value: p  

1

​  

=4,p  

2

​  

=4,p  

3

​  

=1,p  

4

​  

=1

Then,

n=a  

1

4

​  

×a  

2

4

​  

×a  

3

1

​  

×a  

4

1

​  

 

=2  

4

⋅3  

4

⋅5  

1

⋅7  

1

 

=16⋅81⋅5⋅7  

=45360

Hence, the smallest number is 45360