Three is the Problem

If an integer $\displaystyle N = \prod_{k=1}^m p_k^{q_k}$ , where $p_k$ is a prime factor of $N$ , then the number of positive factors of $N$ is given by $\displaystyle n = \prod_{k=1}^m (q_k +1)$ . Since the only way to get $n=3$ is $n = 2+1$ , $N$ must be of the form $N=p^2$ or square of a prime $p$ . For $100 \le N \le 999$ , $10 \le p \le 31$ and they are $p = 11, 13, 17, 19, 23, 29, 31$ , $\boxed{7}$ of them.

Bonus: For $n=5$ , again the only possible case is $n=4+1$ ; and $N = p^4$ . For $10000 \le N \le 99999$ , $10 \le p \le 17$ . That is $p=11, 13, 17$ . 3 solutions only.

Any number with an odd number of factors must be a square, because otherwise each factor has a complement that it is multiplied with to get the number.

Moreover, its square root will have exactly 3 factors. Which implies that the square root itself must be a perfect square.

Since the square root has 3 factors, square root of square root will have exactly two factors. Only prime numbers have exactly two factors.

So if a number has exactly five factors then it must be of the form p^4, where p is a prime number.

In order for p^4 to be a five digit number, p must be between 10 and 18. Hence, p = 11,13,17

There are three such numbers.