$\lfloor \sqrt{n-100} \rfloor$ dividing $n$ 
For how many positive integers $100 < n \le 10000$ does $\lfloor \sqrt{n-100} \rfloor$ divide $n$?

Let $n = 100+k^2+c$, where $k$ is a nonnegative integer and $c$ an integer such that $0 \leq c < 2k+1$. Then $k = \lfloor \sqrt{n-100} \rfloor$ and $1 \leq k \leq 99$ and we have $$\dfrac{n}{\lfloor \sqrt{n-100} \rfloor} = \dfrac{100}{k}+\dfrac{c}{k}+k.$$
How do we continue from here?

First, fix $k = \lfloor \sqrt{n-100} \rfloor \in [1,98]$. Then, $100+k^2 \le n \le 100+k^2+2k$.
Exactly two of the integers $\{100+k^2+c : c \in \mathbb{N}, 1 \le c \le 2k\}$ will be divisible by $k$ (since these are $2k$ consecutive integers).
Also, $100+k^2$ will be divisible by $k$ iff $100$ is divisible by $k$, which happens exactly for $k = 1,2,4,5,10,20,25,50$ (a total of $8$ values of $k$).
Therefore, for $k = 1,2,4,5,10,20,25,50$, there are $3$ integers of the form $n = 100+k^2+c$ where $0 \le c \le 2k$ that are divisible by $k$, and for all other values of $k \in [1,98]$ (a total of $98-8 = 90$ values of $k$), there are $2$ such integers. 
Finally, for $k = 99$, we must have $99^2+100 = 9901 \le n \le 10000$. The only value of $n$ in this range which is divisible by $k = 99$ is $9999$. 
Thus, the total number of integers $100 < n \le 10000$ such that $\lfloor \sqrt{n-100} \rfloor$ divides $n$ is $8 \cdot 3 + 90 \cdot 2 + 1 = 205$.
