Knowledge required: number theory, binary search

First of all, it is important to recognize the relationship between $N!$ and its number of trailing zeros. A zero is formed with a pair of a factor of $2$ and a factor of $5$. In fact, it is not necessary to account for the count of both, since the number of zeros is dependent on the lesser of the two, and the number of $5$ will always be less or equal than that of $2$s.

To solve for the 50% subtask, we can simply brute-force the value of $N$ such that $N!$ will have a number of trailing zeros in the range of $[a,b]$.

Time Complexity: $\mathcal{O}(b)$

To solve for the rest of the cases, it is important to recognize that the number of trailing zeros of $N!$ will always be non-decreasing in relationship to an increasing $N$. Hence, we can use binary search to find:

1. The minimum $N$ such that $N!$ has at least $a$ trailing zeros.
2. The maximum $N$ such that $N!$ does not have more than $b$ trailing zeros.

The number of valid values of $N$ can be calculated by subtracting the two and adding one.

Time Complexity: $\mathcal{O}({\log }_{2}max(b))$