Any number can have only $3$ factors when it is the square of a prime.

The range for $3$ -digit prime-squares is from $11 (121)$ to $31 (961)$ .

The numbers of of primes between $11$ and $31$ is $\boxed{7}-11, 13,17,19,23,29,31$

If a number greater than 1 is NOT a power of a prime, then it has at least 4 different divisors. In fact if $p,q$ are prime numbers such that $p < q$ and both are divisors of a number $n$ so we have $1, p, q, pq$ are four different divisors of $n$ . So the number IS a power of a prime. It is very easy now to see that $p^n$ has exactly $n+1$ divisors, namely

$1=p^0, p=p^1, \ldots , p^n$

Since the number has 3 divisors and is a power of a prime, then it must have the form $p^2$ . It is a prime squared.

Any number having three factors has to be a square of a prime no. so tht factorising that number would give only 1, the number itself and the square root of that no. So there are only 7 such no. between 100 & 999(where prime no. are decided as 6n+/_ 1; where n=1,2....)

• Any prime has 2 factors by definition
• A product pq of 2 primes p and q has factors: $\{1,p,q,pq\}$ . This set contains 3 members iff p=q.
• Any product pqr... of 3 or more primes will have more than 3 factors : $\{1,p,pq,pqr,... \}$

So, only prime squares qualify. The primes must be in the range $\sqrt{100} \leq p \lt \sqrt{1000}$ so that our prime squares are $11^2,13^2,17^2,19^2,23^2,29^2,31^2$

Here's a little Ruby script that answers the question:

1

[*100..999].count{ |x| (1..x).count { |i| x % i  == 0 } == 3 }

Only squares can have an odd number of factors (every factor has a counterpart that is different except squares as the factor is repeated) For example 12 has (1,12) (2,6) (3,4) but 16 has (1,16) (2,8) (4,4) Any square with non prime factors also has the factors of those numbers as well, 8 is a factor of 16 so automatically 2 and 4 are as well. This does not happen with prime factors. 9 for example only has 1, 3 and 9, because 3 has no factors other than 1 and 3

Let one of the 3-digit numbers that we are looking for be $N$ for $100 \le N \le 999$ . It should be noted that $N$ can be represented in the form of primes factorization i.e $N =p_{1}^{k_1}p_{2}^{k_2}...p_{j}^{k_j}$ for primes $p_{1},p_{2},...,p_{j}$ and natural numbers $k_{1},k_{2},...,k_{j}$

The total amount of positive divisor of $N$ ( denoted $\tau(N)$ ) is $(k_{1}+1)(k_{2}+1)...(k_{j}+1)$

We know that $\tau(N) = (k_{1}+1)(k_{2}+1)...(k_{j}+1) = 3$

Since $k_{1},k_{2},...,k_{j}$ are all natural numbers and is by definition; does not include zero, then the only value for $\tau(N) = 3$ is when $j =1$ . It's impossible to have any $j \ge 2$ because if so, an example would be; $\tau(N) = (k_{1}+1)(k_{2}+1) = 3$ . In this case, either of the $k_{1},k_{2}$ could be equal to $0$ while the other one could be equal to $2$ , but we know that $k_{j}$ must be a natural number. Therefore, having $j \ge 2$ is impossible, implying that $j =1$

Now:

$\tau(N) = k_{1}+1 = 3 \Leftrightarrow k_{1} = 2$

Recalling $N = p_{1}^{k_1} = p^{2}$

This suggests that the particular 3-digit number $N$ that we are finding, must be a perfect square with the base of this square being a prime, that ranges from $100 \rightarrow 999$ . This gives us all values of $p^2$ being from $p=11$ to $p=31$ , and between $11 \rightarrow 31$ inclusive, there are $7$ primes. Therefore, there are exactly $\boxed{7}$ 3-digits number that have exactly 3 factors