Number of integers satisfying $\left[\frac{x}{100}\left[\frac{x}{100}\right]\right]=5$ Find Number of integers  satisfying $$\left[\frac{x}{100}\left[\frac{x}{100}\right]\right]=5$$ where $[.]$ is Floor function.
I assumed $$x=100q+r$$ where $0 \le r \le 99$
Then we have 
$$\left[\left(q+\frac{r}{100}\right)q\right]=5$$ $\implies$
$$q^2+\left[\frac{rq}{100}\right]=5$$
Since $rq$ is an integer we have $$rq=100p+r_1$$ where $0 \le r_1 \le 99$
Then we have
$$q^2+p+\left[\frac{r_1}{100}\right]=5$$ $\implies$
$$q^2+p=5$$ so the possible ordered pairs $(p,q)$ are
$(1,2)$, $(1,-2)$, $(-4, 3)$ i am getting infinite pairs.
How to proceed?
 A: Using
$$ [x]\le x<[x]+1 $$
one has
$$ 5\le\frac{x}{100}\left[\frac{x}{100}\right]<6. \tag{1}$$
Let $x=100q+r$, where $q$ and $r$ are an integer and $r\in[0,99]$. Then (1) becomes
$$ 500\le (100q+r)q<600. \tag{2}$$
Noting that
$$ 100q^2 \le (100q+r)q<600$$
implies $q^2<6$, one has $q=\pm1,\pm2$. If $q=\pm1$, it is easy to see that (2) can't hold.
So $q=\pm2$. If $q=-2$, (2) becomes
$$500\le-2(-200+ r)<600 $$
which gives
$$ 50\le -2r<100$$
which is impossible since $r\ge0$.
If $q=2$, (2) becomes
$$500\le2(200+ r)<600 $$
which gives
$$ 50\le r<100.$$
Thus $r=50,51,\cdots99$ and hence
$$ x=250,251,\cdots,299. $$
A: A less-than-satisfying, but simple and fast computational approach -- this double-flooring operation is increasing in $x$. When $x=200$, you get 4. When $x=300$, to get 9. This just leaves 100 integers to check with a simple program. Here is one such program in Mathematica:

Tally@Table[Floor[x/100*Floor[x/100]], {x, 200, 300}]

Which yields

{{4, 50}, {5, 50}, {9, 1}}

[edit] A quick edit -- if you're also concerned with negative integers, a similar approach can be taken to see that we need only check between $x=-300$ and $x=-200$ to get all of those too.
A: I got $x\in [|250,299|]$ 
I solved it writing $x = a10^2+b10^1+c10^0$. And reasoning on $a$ then $b$ then $c$ given the fact that $f$ is non decreasing. 
$f(200) = 4$ and
$f(300) = 9$
Therefore $a = 2$

$f(240) = 4$ and 
$f(250) = 5$ and
$f(290) = 5$
Therefore $5\leq b \leq 9$

$f(299) = 5$
Therefore $0\leq c\leq 9$
