$7^n | \binom{2016}{1003}$

Question

$7^n | \binom{2016}{1003}$. Find the largest value of $n$.

I am a newbie in number theory. Only the thing I know about this is $$\binom{2016}{1003} = \frac{2016!}{1003! \times 1013!}$$
How to proceed ?

Answer 1

Let, $n!= 7^x \times \{something\}$ then $x$ is given by Legendre's_formula - $$L_7(n) = \sum_{i=1}^{\infty} \big{\lfloor}\frac{n}{7^i} \big{\rfloor} $$
Now Let, $$\frac{2016!}{1003!\times 1013!} = \frac{7^a \times {C_1}}{7^b \times C_2 * 7^c \times C_3} = 7^{a-(b+c)} \times {something}$$
So, the largest power of $7$ = $L_7(2016) - L_7(1003) - L_7(1013)$
You will never need $7^4$ as $7^4 > 2016,1003,1013$. SO you can solve it in hand.

Answer 2

The exponent $v_p$ of a prime $p$ in a factorial $n!$ (i.e., the unique integer with $p^{v_p}\mid n!$ and $p^{v_p+1}\nmid n!$) can be found as $$ v_p=\left\lfloor \frac np\right\rfloor +\left\lfloor \frac n{p^2}\right\rfloor+\left\lfloor \frac n{p^3}\right\rfloor+\ldots$$ (where only finitely many summands need to be considered because $\left\lfloor \frac n{p^k}\right\rfloor=0$ as soon as $p^k>n$).

Answer 3

In base seven we have $2016=5610_7$, $1013=2645_7$ and $1003=2632_7$. So when you add $1003+1013$ in base seven there will be three carries (at positions 1,2,3). Therefore, by Kummer's theorem, the answer is $7^3$.