Largest 3-digit prime factor

note that $\binom{2015}{1007} = \frac{2015!}{1007!1008!}$

let $p$ is prime factor of $\binom{2015}{1007}$

imagine $V_{p} \left ( n! \right )$ is the largest value of $k$ such that $p^{k} .m = n!$ with $GCD(p,m)=1$ and $p,k,m,n \in \mathbb{N}$

we know

$V_{p} \left ( 2015! \right ) = \sum_{k=1}^{\infty} \left \lfloor \frac{2015}{p^{k}} \right \rfloor$ $V_{p} \left ( 1008! \right ) = \sum_{k=1}^{\infty} \left \lfloor \frac{1008}{p^{k}} \right \rfloor$ $V_{p} \left ( 1007! \right ) = \sum_{k=1}^{\infty} \left \lfloor \frac{1007}{p^{k}} \right \rfloor$
$V_{p} \left ( \binom{2015}{1007} \right ) = V_{p} \left ( 2015! \right ) - V_{p} \left ( 1008! \right ) - V_{p} \left ( 1007! \right )$

so we have $V_{p} \left ( \binom{2015}{1007} \right ) = \sum_{k=1}^{\infty} \left \lfloor \frac{2015}{p^{k}} \right \rfloor - \sum_{k=1}^{\infty} \left \lfloor \frac{1008}{p^{k}} \right \rfloor - \sum_{k=1}^{\infty} \left \lfloor \frac{1007}{p^{k}} \right \rfloor$

WLOG for $p$ maximum we need to simplify that sum with only take $k =1$

then $V_{p} \left ( \binom{2015}{1007} \right ) = \left \lfloor \frac{2015}{p} \right \rfloor - \left \lfloor \frac{1008}{p} \right \rfloor - \left \lfloor \frac{1007}{p} \right \rfloor$

$\,\,\,\,\,$ the other way to say that when $p$ is maximum $V_{p}$ is minimum.

WLOG let $V_{p} = 1$

So $\left \lfloor \frac{2015}{p} \right \rfloor - \left \lfloor \frac{1008}{p} \right \rfloor - \left \lfloor \frac{1007}{p} \right \rfloor = 1$

Try all case posible but WLOG again for

$\left \lfloor \frac{2015}{p} \right \rfloor = 3$ then $\left \lfloor \frac{2015}{4} \right \rfloor < p \leq \left \lfloor \frac{2015}{3} \right \rfloor \rightarrow \, 503 < p \leq 671$ $$

$\left \lfloor \frac{1008}{p} \right \rfloor = 1$ then $\left \lfloor \frac{1008}{2} \right \rfloor < p \leq 1008 \rightarrow \, 504< p \leq 1008$ $$

$\left \lfloor \frac{1007}{p} \right \rfloor = 1$ then $\left \lfloor \frac{1007}{2} \right \rfloor < p \leq 1007 \rightarrow \, 503 < p \leq 1007$ $$

So interval $p$ is $505 \leq p \leq 671$

$p$ is prime , then the Largest prime for $p$ is $p_{max} = \boxed{661}$

When written out, the number would be: (2015x2014x...x1009)/(1007!). As the answer has to be prime, its multiples have to be the one to determine whether it is divisible or not. Starting at 999, you will notice that the denominator has a 999, and the numerator will have a 999x2, which will cancel out. Therefore, for a number to be a factor of the large number, there will have to be more of it above than below. That leads you to this formula: 3x<2015. This means that there has to be 3 multiples of the answer before 2015, so that there can be 2 in the numerator and 1 in the denominator. This leads to x<671.6666..., and the largest prime number smaller than this is 661.