#5 Measure your Calibre

$\lfloor\frac{x}{100}\lfloor\frac{x}{100}\rfloor\rfloor=5$

If $\frac{x}{100}=t$

$\lfloor t\lfloor t \rfloor \rfloor=5$

$5 \leq t\lfloor t \rfloor < 6$ ...(1)

$(t-1) < \lfloor t \rfloor \leq t$

Multiplying t,

$t(t-1) < t\lfloor t \rfloor \leq t^2$

Combining the 2 inequalities, we get

$t(t-1) < t\lfloor t \rfloor <6$ And $5 \leq t\lfloor t \rfloor \leq t^2$

So we can see that $5 \leq t^2$ And $t(t-1)<6$ .

$2.23 < |t|$ And $-2<t<3$

We can see that only solutions will be between $2.23<t<3$

So put $\lfloor t \rfloor=2$ in (1)

$5 \leq t*2 < 6$

$2.5 \leq t < 3$

$\frac{x}{100}=t$ So

$250 \leq x < 300$

$x \in (250,251,...,299$

*50 numbers *

$\begin{aligned} \left \lfloor \frac x{100} \left \lfloor \frac x{100} \right \rfloor \right \rfloor & = 5 & \small \color{#3D99F6} \text{Let }u = \frac x{100} \\ \left \lfloor u \left \lfloor u \right \rfloor \right \rfloor & = 5 & \small \color{#3D99F6} \text{Note that }u = \lfloor u \rfloor + \{u\} \\ \left \lfloor (\lfloor u \rfloor + \{u\}) \left \lfloor u \right \rfloor \right \rfloor & = 5 \\ \lfloor u \rfloor^2 + \left \lfloor \lfloor u \rfloor \{u\} \right \rfloor & = 5 \end{aligned}$

Since $0 \le \{x\} < 1$ .

$\lfloor u \rfloor^2 \le 5 \le \lfloor u \rfloor^2 + \lfloor u \rfloor - 1$ $\implies \begin{cases} - 2 \le \lfloor u \rfloor \le 2 \\ \lfloor u \rfloor^2 + \lfloor u \rfloor - 6 \ge 0 & \implies \lfloor u \rfloor \le - 3 \cup \lfloor u \rfloor \ge 2 \end{cases}$ $\implies \lfloor u \rfloor = 2$ . When $\lfloor u \rfloor = 2$ ,

$\begin{aligned} 2^2 + \left \lfloor 2 \{u\} \right \rfloor & = 5 \\ \left \lfloor 2 \{u\} \right \rfloor & = 1 \\ \implies 0.5 \le \{u\} & < 1 \\ 2.5 \le u & < 3 & \small \color{#3D99F6} \text{Note that }100u = x \\ 250 \le x & \le 299 & \small \color{#3D99F6} \text{For integer values} \end{aligned}$

Therefore, there are $\boxed{50}$ integer solutions.