6. The product $1!\cdot 2!\cdot 3!\cdot \cdots \cdot 99!\cdot 100!$ ends with $\qquad$ consecutive 0s.
(Hong Kong, China, provided)
6. 1124.

Since in the product $1!\cdot 2!\cdot 3!\cdots \cdots 99!\cdot 100!$, there are a large number of factor 2s, the number of consecutive 0s at the end is determined by the number of factor 5s. The number of factor 5s in each factorial $x$! is shown in Table 1.
Thus, the total number of factor 5s is
$$
\begin{array}{l}
0+5(1+2+3+4+6+7+\cdots+10+ \\
12+13+\cdots+16+18+19+\cdots+22)+24 \\
=5\left(\frac{22 \times 23}{2}-5-11-17\right)+24 \\
=1124 .
\end{array}
$$

Therefore, the required value is 1124.