<s>[INST] How many positive integers from 1-1000 have 5 divisors? 
How many positive integers from $1-1000$ have $5$ divisors?

Any answers would be greatly appreciated. If you have any questions, I will edit for clarification.
 [/INST] (Filling in some details, on the off chance that other answers are unfamiliar to potential readers!) 
Consider the prime factorization of $75 = 3^1 \cdot 5^2$. How many factors does $75$ have? Well, in forming a factor of $75$, we have to choose how many $3$s to include - there is one available, but we could also choose none; so, the number of choices is $1+1=2$ - and how many $5$s to include - there are two available, but we could also choose none; so, the number of choices is $2+1=3$. The aforementioned $2$ choices and $3$ choices yield $2 \cdot 3 = 6$ combinations using the prime factors, and these are all the factors of $75$: 
$$3^0 \cdot 5^0, 3^0 \cdot 5^1, 3^0 \cdot 5^2, 3^1 \cdot 5^0, 3^1 \cdot 5^1, 3^1 \cdot 5^2$$
More generally, a number with prime factorization $p^a \cdot q^b$ for distinct primes $p$ and $q$ has $(a+1)(b+1)$ factors, since you can form these factors by making one of $a+1$ choices for how many $p$s to include (ranging from $0$ to all $a$) and one of $b+1$ choices for how many $q$s to include (ranging from $0$ to all $b$). Even more generally, you arrive at the formula in Bernard's answer.
So: In one's investigation of $5$, it becomes clear that this can arise as the number of factors if, and only if, it came from a number of the form $p^4$ for a prime $p$. All that remains now is to observe how many primes to the fourth power are present between $1$ and $1000$; this is precisely what is done in Mohammad Riazi-Kermani's answer: just $2^4 = 16$, $3^4 = 81$, and $5^4 = 625$. Already we have $6^4 = 1296 > 1000$, so the three listed numbers are exhaustive.
As a follow up exercise, you may wish to count how many factors $4^4$ and $6^4$ have. 
(Which would you guess has more? Etc.)
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