3. Determine all natural numbers $n$ and $k, k>1$, such that $k$ divides each of the numbers

$$
\binom{n}{1},\binom{n}{2}, \ldots,\binom{n}{n-1}
$$
3. Lemma (Kummer's Theorem). The exponent of a prime number $p$ in the binomial coefficient $\binom{n}{m}(0 \leqslant m \leqslant n)$ is equal to the number of digit carries in base $p$ when adding $m$ and $n-m$.

Proof. The exponent of the number $p$ in $\binom{n}{m}=\frac{n!}{m!(n-m)!}$ is $\sum_{i>0}\left(\left[\frac{n}{p^{2}}\right]-\left[\frac{m}{p^{i}}\right]-\right.$ $\left.\left[\frac{n-m}{p^{2}}\right]\right)$. In this sum, the $i$-th term is equal to 1 if there is a digit carry at $p^{i-1}$, and 0 otherwise. This implies the statement.

From the condition, it follows that $k \mid n$. Consider some prime divisor $p$ of the number $k$; let $p^{r} \| n$. Since the lemma implies that $p \nmid\binom{n}{p^{r}}$ (there is no digit carry), the only possibility is $n=p^{r}$; hence, $k$ is also a power of $p$. The lemma also implies $p^{1} \|\binom{n}{p^{k-1}}$, so it must be that $k=p$.