If you don't know how to handle a problem, you should try to look at a few examples first (and in this case, see how the hint plays a role in the example).
Let's take a look at the problem for . We'd like to count up the number of divisors for . We can just use the "elementary school" method, by looking at pairs of divisors (we get one "big" and one "small): and , and , and , and so on. When can we stop checking? We stop at , since and its corresponding "big" divisor would exceed . (In other words, they would have to multiply to or more.) So with respect to the original problem, if we want to count up the divisors, we only need to find the divisors less that and double them (since everything comes in pairs). In this case, there are divisors, which is less that .
Can you see how to apply this for the general ?