Example 1 (9th American Invitational Mathematics Examination AIME Problem) Let $r$ be a real number satisfying the condition:
$$
\left[r+\frac{19}{100}\right]+\left[r+\frac{20}{100}\right]+\left[r+\frac{21}{100}\right]+\cdots+\left[r+\frac{91}{100}\right]=546 \text {. }
$$

Find $[100 r]$.
Let $[r]=n$, and suppose that $\left[r+\frac{19}{100}\right],\left[r+\frac{20}{100}\right], \cdots,\left[r+\frac{91}{100}\right]$ contains $k$ instances of $n+1$ and $73-k$ instances of $n, (0 \leqslant k \leqslant 73)$.
Thus, according to the problem, we have $(73-k) n+k(n+1)=546$, which simplifies to $n=7+\frac{35-k}{73}$. Therefore, $k=35, n=7$. This gives us $r+\frac{56}{100}<8, r+\frac{57}{100} \geqslant 8$, or $743 \leqslant 100 r<744$. Hence, $[100 r]=743$.