I have this question: n is a positive integer. s is a largest non negative integer such that $2^{s}|n$. How can I prove $2^{s}|\binom{n}{k}$ for every odd integer k, so that $1 \leq k \leq n$.
Well I know that if $1 \leq k \leq n$, then $\frac{n}{gcd(n,k)}|\binom{n}{k}$. This is from Google Searching. But how is it I can use this to prove the equation. I haven't solved it. But can I make $$n=p_{1}^{\alpha_{1}}p_{2}^{\alpha_{2}}...p_{m}^{\alpha_{m}}$$ and $$k=q_{1}^{\beta_{1}}q_{2}^{\beta_{2}}...p_{m}^{\beta_{m}}$$
Then $\gamma_{i} =\text{min(}\alpha_{i}, \beta_{i}\text{)}, i=1,2,....m$.Thank you for your time.
No need to use all those primes, you just care about $2.$
If $n$ is odd, then $s=0$ so there is nothing to show because $1$ divides every number.
If $n$ is even and $k$ is odd, then $n/gcd(n,k)=2^s\cdot n_1,$ where $n_1=\frac{n}{2^sgcd(n,k)},$ notice that $gcd(n,k)$ is odd because it has to divide $k,$ so $n_1$ is an integer and we have that $n/gcd(n,k)=2^s\cdot n_1|\binom{n}{k},$ hence $2^s|\binom{n}{k}$ by transitivity.