3.3 What is the greatest two-digit prime factor of the integer $C_{2000}^{100}$?
(1st American Invitational Mathematics Examination, 1983)
[Solution]
$$
\begin{aligned}
n & =C_{200}^{100} \\
& =\frac{200 \cdot 199 \cdot 198 \cdot \cdots \cdot 102 \cdot 101}{100!} \\
& =2^{50} \cdot \frac{199 \cdot 197 \cdot 195 \cdot \cdots \cdot 103 \cdot 101}{50!} .
\end{aligned}
$$

This is an integer, and the product $199 \cdot 197 \cdot 195 \cdot \cdots \cdot 103 \cdot 101$ in the numerator 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 any odd composite number 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.