Determine the largest 3-digit prime factor of ${2000 \choose 1000}$ Determine the largest 3-digit prime factor of ${2000 \choose 1000}$.  
I could not approach the problem at all. I have no idea how to try the problem. Please help.
 A: Hint: Show that if $p$ is a prime such that $667 \leq p \leq 1000$, then $ p$ is not a factor of $ {2000 \choose 1000}$. Use the fact that $667 \times 3 > 2000$.
Hint: Try the next largest prime number $p^*$. Does it work?
A: To find the solution, I need to find a number that comes exactly thrice in $2000!$ but exactly  once in $1000!$ as, $3-2=1$. We get the $2$ as $1000!$ occurs twice. So, such a number is the prime just below $666$ which is $661$.
A: Kummer's theorem, says that the number of factors $p$ (prime) in a binomial coefficient $\binom{a+b}a$ equals the number of carries produced while performing the addition $a+b$ using base$~p$ representation of integers. Since the question only considers prime factors $p<1000$, the representation of $1000$ in base$~p$ contains at least two digits; moreover if $p>500$ the first digit will be$~1$. Now the only way $1000+1000$ can produce a carry in base$~p$ for $500<p<1000$ is if the (equal) final digits do so, and for that their digit value should be larger than $p/2$. This means that $1000>p+p/2$, which gives $p<\frac23\times1000<667$. Moreover, for values of $p$ close to that upper bound one will have $1000\bmod p=1000-p>p/2$, so it suffices to search for the largest prime number $p<667$.
