Let n be a positive integer with exactly 3 positive prime divisors. If n^2 has 27 divisors, how many does n have?

Guest Sep 24, 2021

#1**+1 **

**n==30**

**30==2 * 3 * 5==3 prime factors**

**30==(1, 2, 3, 5, 6, 10, 15, 30)>>Total = 8 divisors**

**30^2= (1, 2, 3, 4, 5, 6, 9, 10, 12, 15, 18, 20, 25, 30, 36, 45, 50, 60, 75, 90, 100, 150, 180, 225, 300, 450, 900) > 27 (divisors)**

Guest Sep 24, 2021

#2**+2 **

Thanks Guest, you have utilized C++ (I assume) and answered for a specific case but this was reasonable.

I have given a more general answer. But I liked your answer

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Let n be a positive integer with exactly 3 positive prime divisors. If n^2 has 27 divisors, how many does n have?

** All the pronumeral I have used are positive integers. a,b and c are also prime numbers

\(n=a^pb^qc^r\qquad \text{This will have }(p+1)(q+1)(r+1)\;\;factors\\ n^2=a^{2p}b^{2q}c^{2r} \qquad\text{This will have }(2p+1)(2q+1)(2r+1)\;\;factors \)

We are told that

\((2p+1)(2q+1)(2r+1)=27\)

Since p,q and r are all positive integer the smallest value that any of those brackets can take is 3

3^3=27

so that means that p=q=r=1

so n must have 2*2*2=8 factors.

LaTex:

n=a^pb^qc^r\qquad \text{This will have }(p+1)(q+1)(r+1)\;\;factors\\

n^2=a^{2p}b^{2q}c^{2r} \qquad\text{This will have }(2p+1)(2q+1)(2r+1)\;\;factors

Melody Sep 24, 2021