6. Find the value of $r$ such that
$$
\begin{array}{l}
{\left[r+\frac{19}{100}\right]+\left[r+\frac{20}{100}\right]+\cdots} \\
+\left[r+\frac{91}{100}\right]=546 .
\end{array}
$$

Find $[100 r]$. (Where $[x]$ denotes the greatest integer not greater than $x$)
6. The left side of the equation has 73 terms, each being $[r]$ or $[r+1]$. This is because $\frac{19}{100}, \frac{20}{100}, \cdots, \frac{91}{100}$ are all less than 1, and $73 \times 7 < 546 < 73 \times 8$. To make the sum 546, $[r]$ must be 7.

Assume $\left[r+\frac{k}{100}\right]=7(19 \leqslant k \leqslant m),[r+$ $\left.\frac{k}{100}\right]=8(m+1 \leqslant k \leqslant 91)$. Then $7(m-18)+$ $8(91-m)=546$, yielding $m=5.6$, so $\left[r+\frac{56}{1.00}\right]=7$, $\left[r+\frac{5 ?}{100}\right]=8$, thus $7.43 \leqslant r \approx 7.44$. Therefore, $[100 r]=743$.