Find the smallest positive integer $m$ such that $ {2015}\choose{m} $ is an even number

Question

Find the smallest positive integer $m$ such that $$
{2015}\choose{m}$$ is an even number.

Since $$
{{2015}\choose{m}} = \frac{2015}{1} \cdot \frac{2014}{2} \cdots \frac{2016-m}{m} = \prod_{k=1}^m \frac{2016-k}{k}, $$
we only need to find the smallest $m$ such that $$ m = 2^{a_m} \cdot p_m, \, 2016-m = 2^{b_m} \cdot q_m, \, 2 \not \mid p_m, \, 2 \not \mid q_m, \, a_m < b_m.$$

In this problem, it turns out that when $m=32$, we have $$ 32=2^5, \, 2016-32=1984=2^6 \cdot 31. $$

However, we will need to try $m=2,4,6,\ldots,32$, the answer will not come out easily.

Is there any easier way to solve this problem?

Answer 1

Kummer's theorem:

for given integers $n \ge m \ge 0$ and a prime number $p$, the $p$-adic valuation $\nu _{p}\left({\tbinom {n}{m}}\right)$ is equal to the number of carries when $m$ is added to $n - m$ in base $p$

Since $2015_{10} = 11111011111_2$ it's clear that for any $m < 32$ there will be no carries, and so $\binom{2015}{m}$ will be odd. However, to subtract $32_{10} = 100000_2$ we need to borrow, and therefore adding $32$ and $2015-32$ will require a carry. QED.