10. (1983 AIME, Problem 1) What is the greatest two-digit prime factor of the integer $C_{200}^{100}$?
10. $n=C_{200}^{100}=\frac{200 \cdot 199 \cdot 198 \cdot \cdots \cdot 102 \cdot 101}{100!}=2^{50} \cdot \frac{199 \cdot 197 \cdot 195 \cdot \cdots \cdot 103 \cdot 101}{50!}$,

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

Since any composite number less than 200 must have a prime factor less than $\sqrt{200}$, the prime factors of the odd composites between 101 and 199 must include one of $3, 5, 7, 11, 13$.

Since $\left[\frac{199}{3}\right]=66$, 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 $61 \cdot 3=183$ is indeed one of the factors in $C_{200}^{100}$, so the largest prime factor we are looking for is 61.