Show that exists some positive integer $N$ such that among $1,2,...,N$, there are at least $0.99N$ good numbers. Given a positive integer $k$, call $n$ good, if among $$\binom{n}{0},\binom{n}{1},\binom{n}{2},...,\binom{n}{n}$$ at least $0.99n$ of them are divisible by $k$. Show that exists some positive integer $N$ such that among $1,2,...,N$, there are at least $0.99N$ good numbers.
some try: for all prime $p$ and nonnegative integers $m,n$, 
$$\nu_p \binom{m+n}{n} = \frac{s_p(m)+s_p(n)-s_p(m+n)}{p-1}$$
which is equal to the number of carries when adding $m$ and $n$ in mod $p$.
 A: It suffices to prove that for every $p^a$ the set of $p^a$-good numbers has density $1$.
To do this we first find an upper bound for the number of values of the form $\binom{n}{k}$ such that $v_p(\binom{n}{k})$ is less than $a$ (  for a fixed value of $n$)
we only use this estimate for values of $n$ that have at least $Ma$ zeros in base $p$ (here $M$ will be a predetermined large enough value, in terms of $A$). Suppose that $\binom{n}{j}$ is not a multiple of $p^a$, then the number of carries when adding $j$ and $n-j$ is less than $a$. This means that at least $(M-1)a$ of the $Ma$ positions which are zeros in $n$ are zeros both in $j$ and $n-j$. If we initially take a large value of $M$ this will guarantee that less than $0.1$ of the possible values of $j$ are not multiples of $p^a$ and subsequently $n$ is good if $n$ has at least $Ma$ zeros in base $p$.
It is easy to see that the numebrs with at least $Ma$ zeros have asymptotic density $1$.
Finally notice that a finite intersection of sets with asymptotic density $1$ also has asymptotic density $1$.
So if we are given $k=p_1^{a_1}\dots p_r^{a_r}$ we can deduce that the set of $k$-good integers has asymptotic density $1$ since the $p_i^{a_i}$ integers have asymptotic density $1$.
Once we know that the $k$-integers have asymptotic density $1$ it follows that an $n$ exists such that at least $0.99n$ of the inetgers between $1$ and $n$ are good.
A: $N=k^8$ is such a number. 
Let $N=k^8$, and call a number $1\le\ell\le N$ "bad" if $\tbinom{N}{\ell}$ is not divisible by $k$. How can that happen? There must be some prime $p$, a divisor of $k$, such that 
$v_p\left(\tbinom{N}{\ell}\right) < v_p(k)$. Due to $\tbinom{N}{\ell} = \tbinom{N-1}{\ell-1} \cdot \tfrac{k^8}{\ell}$,
this implies
\begin{gather*}
8v_p(k) - v_p(\ell) \le
v_p\left(\tbinom{N}{\ell}\right) \le v_p(k)-1 \\
v_p(\ell) \ge 7v_p(k)+1 \ge 8
\end{gather*}
so $p^8$ must divide $\ell$.
Hence, each bad $1\le\ell\le N$ has a divisor of the form $p^8$ and
$$
\text{number of bad $\ell$ values} \le 
\sum_{\text{$p$ prime}} \left\lfloor\dfrac{N}{p^8}\right\rfloor <
N \sum_{m=2}^\infty \frac{1}{m^8} <
N \left( \frac{1}{2^8} + \int_2^\infty\dfrac{\mathrm{d}x}{x^8} \right) 
= \frac{9}{1792} N < \frac{N}{100}.
$$
