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?

  • 1
    $\begingroup$ $205$ is the answer (computer). The way seems very large. $\endgroup$ – MonsieurGalois Dec 29 '16 at 3:37
  • $\begingroup$ Shouldn't it be $\frac{100}{k}+k+\frac{c}{k}$? $\endgroup$ – rtybase Dec 29 '16 at 3:39

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 $8 \cdot 3 + 90 \cdot 2 + 1 = 205$.

  • $\begingroup$ How did you get the $98 \cdot 2$? $\endgroup$ – user19405892 Dec 29 '16 at 3:45
  • $\begingroup$ ^There are $98$ values of $k \in [1,98]$ with at least $2$ integers $n$ and $8$ values of $k \in [1,98]$ with an "extra" value of $n$. I've edited my answer to make it less confusing. $\endgroup$ – JimmyK4542 Dec 29 '16 at 3:47
  • $\begingroup$ We're trying to count the number of integers between $100+k^2$ and $100+k^2+2k$ that are divisible by $k$. I've already stated that exactly $2$ integers between $100+k^2+1$ and $100+k^2+2k$ are divisible by $k$, so the only one that's left is $100+k^2$. For the $98-8 = 90$ values of $k \in [1,98]$ that are not factors of $100$, we have that $100$ is not divisible by $k$, and thus, $100+k^2$ is not divisible by $k$. Thus, there are only $2$ integers between $100+k^2$ and $100+k^2+2k$ which are divisible by $k$ in that case. $\endgroup$ – JimmyK4542 Dec 29 '16 at 3:58

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