Hermite's identity

First note if $r$ is a whole number, all $73$ terms the same whole number. $8 \times 73 =584$ which is too big by $584-546=38$ so we need the first $38$ terms to be $7$ and the last $35$ terms to be $8$ . Checking my math here: $7\times 38 + 8\times 35 = 546$

The $38^{th}$ term is $\lfloor r + \frac{56}{100} \rfloor$ . Since $100-56=44$ , for this to be the last term that gives $7$ and not $8$ we need $7.43\le r<7.44$ , so $\lfloor 100 \rfloor=\boxed{743}$

Let $S(r)$ be the LHS of the equation. Then $\displaystyle S(r) = \sum_{k=1}^{73} \left \lfloor r + \frac {k+18}{100}\right \rfloor$ for integer $r$ , we have $S(r) = 73r$ , and $S(7)=511$ and $S(8)= 584$ . Therefore, the required $r$ such that $S(r) = 546$ is $7 < r < 8$ and that:

$\begin{aligned} S(r) & = \sum_{k=1}^{73} \left \lfloor r + \frac {k+18}{100}\right \rfloor \\ & = 73\lfloor r \rfloor + \sum_{k=1}^{73} \left \lfloor \{r\} + \frac {k+18}{100}\right \rfloor & \small \color{#3D99F6} \text{where } 0 \le \{r\} < 1 \text{ is the fractional part of }r. \\ 546 & = 511 + \sum_{k=1}^{73} \left \lfloor \{r\} + \frac {k+18}{100}\right \rfloor & \small \color{#3D99F6} \text{for }\lfloor r \rfloor = 7 \end{aligned}$

$\displaystyle \implies \sum_{k=1}^{73} \left \lfloor \{r\} + \frac {k+18}{100}\right \rfloor = 35$ . This means that the last 35 terms of the summation is equal to 1. And the first $73-35=38$ terms are less than 1. Therefore, $\{r\} + \dfrac {38+18}{100} < 1$ $\implies \{r\} < 1 - \dfrac {56}{100} = 0.44$ $\implies r = 7+0.43 = \boxed{7.43}$ .