3. If the real number $r$ satisfies $\left[r+\frac{19}{100}\right]+\left[r+\frac{20}{100}\right]+\cdots+\left[r+\frac{91}{100}\right]=546$, then the value of $[100 r]$ is
A. 740
B. 714
C. 743
D. 717
3. C Since $\frac{19}{100}, \frac{20}{100}, \cdots, \frac{91}{100}$ are all less than 1, each term on the left side of the equation is $[r]$ or $[r]+1$. There are 73 terms on the left side, and $73 \times 7 < 546 < 73 \times 8$, so $[r]=7$. Let $\left[r+\frac{k}{100}\right]=7(19 \leqslant k \leqslant m)$ and $\left[r+\frac{p}{100}\right]=8(m+1 \leqslant p \leqslant 91)$, then $7(m-18)+8(91-m)=546$, which gives $m=56$. Therefore, $\left[r+\frac{56}{100}\right]=7$, meaning $6.44 \leqslant r < 7.44$; and $\left[r+\frac{57}{100}\right]=8$, meaning $7.43 \leqslant r < 8.43$. Hence, $7.43 \leqslant r < 7.44$. So $[100 r]=743$.