# 1983 AIME Problems/Problem 8

## Problem

What is the largest $2$-digit prime factor of the integer $n = {200\choose 100}$?

## Solution

Expanding the binomial coefficient, we get ${200 \choose 100}=\frac{200!}{100!100!}$. Let the required prime be $p$; then $10 \le p < 100$. If $p > 50$, then the factor of $p$ appears twice in the denominator. Thus, we need $p$ to appear as a factor at least three times in the numerator, so $3p<200$. The largest such prime is $\boxed{061}$, which is our answer.

## Solution 2: Clarification of Solution 1

We know that $${200\choose100}=\frac{200!}{100!100!}$$ Since $p<100$, there is at least $1$ factor of $p$ in each of the $100!$ in the denominator. Thus there must be at least $3$ factors of $p$ in the numerator $200!$ for $p$ to be a factor of $n=\frac{200!}{100!100!}$. (Note that here we assume the minimum because as $p$ goes larger in value, the number of factors of $p$ in a number decreases,)

So basically, $p$ is the largest prime number such that $$\left \lfloor\frac{200}{p}\right \rfloor>3$$ Since $p<\frac{200}{3}=66.66...$, the largest prime value for $p$ is $p=\boxed{61}$

~ Nafer