$14 \cdot 39$ If the real number $r$ satisfies
$$
\left[r+\frac{19}{100}\right]+\left[r+\frac{20}{100}\right]+\cdots+\left[r+\frac{91}{100}\right]=546 \text {. }
$$

Find $[100 r]$.
(9th American Invitational Mathematics Examination, 1991)
[Solution] Let $r=[r]+\alpha, \quad 0 \leqslant \alpha<1$.
Then the original expression becomes
$$
73[r]+\left[\alpha+\frac{19}{100}\right]+\left[\alpha+\frac{20}{100}\right]+\cdots+\left[\alpha+\frac{91}{100}\right]=546 \text {. }
$$

Since $\left[\alpha+\frac{i}{100}\right]=0$ or $1, i=19,20, \cdots, 91$, then
$$
0 \leqslant \sum_{i=19}^{91}\left[\alpha+\frac{i}{100}\right]<73 \text {. }
$$

Also,
$$
546=73 \cdot 7+35,
$$

so
$$
[r]=7 \text {. }
$$

And
$$
\left[\alpha+\frac{i}{100}\right](i=19,20, \cdots, 91)
$$

contains 35 ones, thus
$$
\left[\alpha+\frac{57}{100}\right]=\left[\alpha+\frac{58}{100}\right]=\cdots=\left[\alpha+\frac{91}{100}\right]=1 .
$$

Therefore,
$$
\frac{43}{100} \leqslant \alpha<\frac{44}{100} .
$$

Since $r=[r]+\alpha=7+\alpha$, we have
$$
\begin{array}{l}
7+\frac{43}{100} \leqslant r<7+\frac{44}{100} \\
743 \leqslant 100 r<744
\end{array}
$$

Thus,
$$
[100 r]=743
$$