Big Choice, No Backsies

$\begin{pmatrix} 600 \\ 300 \end{pmatrix} = \dfrac {600\times 599\times 598\times ... 301} {1\times 2\times 3\times ... 300} = \dfrac {13^m}{13^n} Q$ ,

where $m$ , $n$ and $Q$ are integers.

If $m > n$ then $\begin{pmatrix} 600 \\ 300 \end{pmatrix}$ is divisible by $13$ .

The power $p_{13} (n)$ of the factor 13 of $n!$ is given by:

$p_{13} (n) = \lfloor \frac {n}{13} \rfloor + \lfloor \frac {n}{13^2} \rfloor + \lfloor \frac {n}{13^3} \rfloor +...$ ,

where $\lfloor \space \space \rfloor$ is the greatest integer function.

Now,

$p_{13} (600) = \lfloor \frac {600}{13} \rfloor + \lfloor \frac {600}{169} \rfloor = 46 + 3 = 49$

$p_{13} (300) = \lfloor \frac {300}{13} \rfloor + \lfloor \frac {300}{169} \rfloor = 23 + 1 = 24$

We note that $m = p_{13} (600) - p_{13} (300) = 49 - 24 = 25$ and $n = p_{13} (300) = 24$ . Since $m > n$ , $\begin{pmatrix} 600 \\ 300 \end{pmatrix}$ is $\boxed {divisible}$ by $13$ .

The largest power of 13 in ${n!}$ (for n is any natural number) is given by: $\displaystyle\sum_{k=1}^{\infty} \lfloor \frac{n}{13^k} \rfloor$ (the upper limit is infinity because we eventually obtain 0 + 0 +...) In this case $\displaystyle\sum_{k=1}^{\infty} \lfloor \frac{600}{13^k} \rfloor = \lfloor \frac{600}{13} \rfloor + \lfloor \frac{600}{13^2} \rfloor = 46 + 3 = 49$ $\displaystyle\sum_{k=1}^{\infty} \lfloor \frac{300}{13^k} \rfloor = \lfloor \frac{300}{13} \rfloor + \lfloor \frac{300}{13^2} \rfloor = 23 + 1 = 24$ Now if ${\phi}$ represents the value of the binomial that doesn't contain powers of 13 then we have: ${600 \choose\ 300} = \frac{13^{49}}{13^{24} \times\ 13^{24}}.\phi = \frac{13^{49}}{13^{48}}.\phi = 13\phi \therefore\ 13| {600 \choose\ 300}$

Use luca's theorem.After writing 600 and 300 in base 13 we find that one digit of 300 exceeds 600. So remainder is 0.

600 / 13 + 600/169 - 2 * (300 / 13 + 300/169) > 0