$\begin{array}{l} \text { Eight, the largest two-digit number factor of the number } \mathrm{n}=\binom{200}{100} \text { is } \\ \text { what? } \quad \text { [Note: }\binom{200}{100} \text { is } \mathrm{C}_{2}^{1} 00 \% \text { ] }\end{array}$
Solve $\mathrm{n}=\mathrm{C}_{200}^{+} \times 0=\frac{200 \cdot 199 \cdot 198 \cdots 102 \cdot 101}{100!}$
$$
=2^{50} \cdot \frac{199 \cdot 197 \cdot 195 \cdots 103 \cdot 101}{50!} .
$$

This number is an integer, and the numerator $199 \cdot 197 \cdot 195 \cdots 103 \cdot 101$
is the product of three-digit odd numbers, so the largest two-digit prime factor we are looking for can only be found among the three-digit composite numbers.

Since any composite number less than 200 must have a prime factor less than $\sqrt{200}$, the prime factors of the odd composite numbers between 101 and 199 must include one of $3, 5, 7, 11, 13$. And $\frac{199}{3}$ $=66 \frac{1}{3}$, so the largest two-digit prime factor we are looking for must be less than 66, and the largest prime number less than 66 is 61, and $61 \times 3=183$ is exactly one of the factors in the numerator of $\mathrm{C}_{200}^{100}$, so the largest prime factor we are looking for is 61.