uniformity of combination Let $m>1$, $n$, $k$ be positive integers. Does there exist a positive integer $l>m-1$ such that $\binom{l}{m}  \equiv k \pmod{2^n}$? I see it at princeton companion to mathematics.
 A: Let $\nu_2$ denote the $2$-adic valuation.

Lemma For every $m>1$ there exists $u$ such that
  
  
*
  
*$1\leq u\leq m$;
  
*$\nu_2(u)\geq\nu_2(v)$ for every $1\leq v\leq m$;
  
*if $1\leq v\leq m$ and $\nu_2(u)=\nu_2(v)$ then $v=u$.
  

proof. Clearly there exists $u$ satisfying (1) and (2).
To prove (3) assume $1\leq u<v\leq m$ such that $\nu_2(u)=\nu_2(v)$.
Then $1\leq v-u\leq m$ and $\nu_2(v-u)>\nu_2(u)$ - a contradiction.
Now let $m$ and $u$ as in the lemma.
Then
\begin{align}
p(x)
&=\binom{2^ux-1}m\\
&=\frac 1{m!}\prod_{v=1}^m(2^ux-v)\\
&=\frac{2^{\nu_2(m!)}}{m!}\prod_{v=1}^m(2^{u-\nu_2(v)}x-2^{-\nu_2(v)}v)\\
&=\frac 1d\prod_{v=1}^m(2^{u-\nu_2(v)}x-d_v)
\end{align}
where $d$ and $d_v$ are odd numbers satisfying $m!=2^{\nu_2(m!)}d$ and $v=2^{\nu_2(v)}d_v$.
Consequently, $p$ is a polynomial with coefficients in $\Bbb{\hat Z}_2$, the ring of $2$-adic integers.
Moreover $p(x)\equiv x-d_u\pmod 2$, hence the equation $p(x)\equiv k\pmod 2$ has a solution for every $k$.
Moreover,
$$p'(x)\equiv 1\pmod 2$$
hence the equation $p(x)\equiv k\pmod 2$ has simple roots.
By Hensel's lemma, the simple root of $p(x)\equiv k\pmod 2$ can be lifted to a root of $p(x)=k$ in $\Bbb{\hat Z}_2$.
Consequently, the congruence $\binom lm\equiv k\pmod{2^n}$ has solution for every $n>0$.
