Find the smallest positive integer $m$ such that $ {2015}\choose{m} $ is an even number 
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?
 A: 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.
A: for any natural number $n$, let $v_2(n)$ denote the order to which $2$ divides $n$.
It is not difficult to show that $$v_2(n!)=\sum_{i=1}^{\infty}\Big \lfloor \frac n{2^i} \Big \rfloor\implies v_2(2015!)=2005$$
It follows that we are asking for the least $k$ such that $$v_2(k!)+v_2((2015-k)!)<2005$$
In searching for such a $k$ it is helpful to note that $2^6\,|\,1984$ and that this is the closest integer less than $2015$ which is divisible by a large power of $2$.  Thus it is reasonable to imagine that we want $k$ such that $2015-k=1983$, so $k=32$.  
Note:  This isn't a complete proof, just a strong heuristic to suggest that $32$ is correct.  Given the above, $32$ is the first number I would try...though there is still some work involved in proving that it is minimal.  The inequality shows an easy way to perform the necessary check without heavy computations.
A: $n = 2015_{10} = 11111011111_2$ and from Luca's theorem:
$${2015 \choose m} = \prod_{i=0}^{10} {n_i \choose m_i} \pmod 2$$
where all factors are always $1$ except for ${0 \choose m_5}$ which is $0$ if and only if $m_5 = 1$.
The smallest number having $m_5 = 1$ is $32$.
Generally speaking ${n \choose m}$ is odd if and only if $m \land n = m$, and even otherwise, where $\land$ is meant here as a bitwise operation on binary digits. In other words ${n \choose m}$ is odd if and only if all $1$s in $m$ are $1$ also in $n$.
