Finding the sum of 3 numbers with 5 positive integer divisors. 
What is the sum of the three numbers less that 1000 that have exactly
  five positive integer divisors?

So I see that first since there are only 5 multiples, the number must be a perfect square. Then, I took a look at a few examples. 
$\#1\text{:}$
$$2^2=4$$
$$\implies\text{1, 4}$$
$$\implies\text{2, 2}$$ 
4 then has 3 positive integer divisors. 
$\#2\text{:}$
$$4^2=16$$
$$\implies\text{1, 16}$$
$$\implies\text{2, 8}$$
$$\implies\text{4, 4}$$ 
This means that the square of a square number has 5 positive integer divisors. There are many of these under 1000, so how come there are three? I am probably missing a point here. What is wrong with my logic? How can I solve this problem?
 A: Recall: The number of divisors of a number $n$ whose prime decomposition is
$$n=p_1^{\alpha_1}\cdots p_k^{\alpha_k}$$
Is $(\alpha_1+1)\cdots(\alpha_k+1)$ (Exercise: prove this statement), then, you want this number to be $5$ and each factor being positive you must have all to be one except for one of them that must be $5$. 
$5$ is prime you can only express it as $1\cdot 5$; so if every factor is one, that means the exponent of that prime is $0$ in the factorization of $n$, and the factor that is $5$ tells us there's only one prime in the factorization and that the power is $5-1=4$ (solving for $\alpha_\text{something}$).
With a little algebra you get that $n=p^4$ for some prime number $p$, then, the only prime numbers whose fourth power is less than $1000$ are $p=2,3,5$ so the sum of the four numbers is
$$2^4+3^4+5^4=722$$
A: The three numbers less than 1000 with exactly 5 factors are:


*

*16 {1, 2, 4, 8, 16}

*81 {1, 3, 9, 27, 81}

*625 {1, 5, 25, 125, 625}


So the sum is:  722, just as Max wrote.
