6. If the real number $x$ satisfies $\left[x+\frac{19}{100}\right]+\left[x+\frac{20}{100}\right]+\cdots+\left[x+\frac{91}{100}\right]=546$, then the value of $[100 x]$ is
6. 743 Analysis: Each term on the left is $[x]$ or $[x]+1$. Noting that $73 \times 7 < 546 < 73 \times 8$, we have $[x]=7$. There are 73 terms on the left. Suppose $\left[x+\frac{k}{100}\right]=7(19 \leqslant k \leqslant m),\left[x+\frac{p}{100}\right]=8(m+1 \leqslant p \leqslant 91)$, then $7(m-18)+8(91-m)=546$, so $m=56$. Therefore, $\left[x+\frac{56}{100}\right]=7$, which means $6.44 \leqslant x < 7.44$, and $\left[x+\frac{57}{100}\right]=8$, which means $7.43 \leqslant x < 8.43$. Thus, $7.43 \leqslant x < 7.44$, so $[100 x]=743$.