2. Find the integer $m$ so that
$$
10^{m}<\frac{1}{2} \times \frac{3}{4} \times \frac{5}{6} \times \ldots \frac{99}{100}<10^{m+1}
$$
2. Let $a=\frac{1}{2} \times \frac{3}{4} \times \frac{5}{6} \times \cdots \times \frac{99}{100}=\frac{3}{2} \times \frac{5}{4} \times \frac{7}{6} \times \cdots \times \frac{99}{98} \times \frac{1}{100}$. Hence, $a>\frac{1}{100}=10^{-2}$. Thus, $m \geq-2$.
Now, let $b=\frac{2}{3} \times \frac{4}{5} \times \frac{6}{7} \times \cdots \times \frac{96}{97} \times \frac{98}{99}$. Notice that $\frac{2}{3}>\frac{1}{2}, \frac{4}{5}>\frac{3}{4}, \ldots, \frac{98}{99}>\frac{97}{98}$. Also, since $\frac{99}{100}0$ then $a^{2}<a b$. But $a b=\frac{1}{100}$, so that $a^{2}<\frac{1}{100}$. Hence, $a<\frac{1}{10}=10^{-1}$. Thus, $m \leq-2$. Therefore, $m=-2$.