What is the highest 3 digit prime factor of ${2000 \choose 1000}$?

[kaliprasad]

find the highest 3 digit prime factor of ${2000 \choose 1000}$

 

[kaliprasad]

find the highest 3 digit prime factor of ${2000 \choose 1000}$

Spoiler

${2000 \choose 1000} = \frac{2000!}{1000!1000!}$
so the prime p occurs $ \lfloor \frac{2000}{p} \rfloor - 2\lfloor \frac{1000}{p} \rfloor $ times
now if p is 3 digit $> \frac{2000}{3}$ or $>666$ then
$ \lfloor \frac{2000}{p} \rfloor = 2 $
$ \lfloor \frac{1000}{p} \rfloor = 1 $
so $ \lfloor \frac{2000}{p} \rfloor - 2\lfloor \frac{1000}{p} \rfloor =0 $
if is $< 666$ and $> 500$
$ \lfloor \frac{2000}{p} \rfloor - 2\lfloor \frac{1000}{p} \rfloor >= 1 $
so largest p is largest prime $< 666$ and it is 661.