6. Given a positive integer $k$ that satisfies for any positive integer $n$, the smallest prime factor of $n^{2}+n-k$ is no less than 11. Then, $k_{\text {min }}$ $=$ . $\qquad$
6.43 .

Notice,
$$
\begin{array}{l}
n^{2}+n \equiv 0(\bmod 2), \\
n^{2}+n \equiv 0,2(\bmod 3),
\end{array}
$$
$$
\begin{array}{l}
n^{2}+n \equiv 0,2,1(\bmod 5), \\
n^{2}+n \equiv 0,2,6,5(\bmod 7),
\end{array}
$$

and the smallest prime factor of $n^{2}+n-k$ is not less than 11, then
$$
\begin{aligned}
k & \equiv 1(\bmod 2), \quad k \equiv 1(\bmod 3), \\
k & \equiv 3,4(\bmod 5), k \equiv 1,3,4(\bmod 7) .
\end{aligned}
$$

Therefore, $k \equiv 43,73,109,169,193,199(\bmod 210)$.
Thus, $k_{\min }=43$.