7.1. Let $a_{1}=1, a_{2}=3$ and $a_{n+2}=(n+3) a_{n+1}-(n+2) a_{n}$ for every integer $n \geq 1$. Find all values of $n$ for which 11 divides $a_{n}$.
7.1. By direct calculation we get

$$
\begin{aligned}
a_{1} & \equiv 1(\bmod 11) \\
a_{2} & \equiv 3(\bmod 11) \\
a_{3} & \equiv 9(\bmod 11) \\
a_{4} & \equiv 0(\bmod 11) \\
a_{5} & \equiv 10(\bmod 11) \\
a_{6} & \equiv 4(\bmod 11) \\
a_{7} & \equiv 6(\bmod 11) \\
a_{8} & \equiv 0(\bmod 11) \\
a_{9} & \equiv 1(\bmod 11) \\
a_{10} & \equiv 0(\bmod 11) \\
a_{11} & \equiv 0(\bmod 11)
\end{aligned}
$$

So from the relation in the problem we obtain that for every $n \geq 10$ we have $a_{n+2} \equiv 0(\bmod 11)$. The answer is:

$$
n \in\{4,8\} \cup\{n \in \mathbf{N}: n \geq 10\}
$$

Alternate solution. From the recurrence relation in the problem we immediately see that for $n \geq 3$ we have

$$
\begin{aligned}
& a_{n}-a_{n-1}=n\left(a_{n-1}-a_{n-2}\right) \\
& a_{n-1}-a_{n-2}=(n-1)\left(a_{n-2}-a_{n-3}\right) \\
& \vdots \\
& a_{4}-a_{3}=4\left(a_{3}-a_{2}\right) \\
& a_{3}-a_{2}=\left(a_{2}-a_{1}\right)
\end{aligned}
$$

After consecutive substitutions we arrive to:

$$
a_{n}-a_{n-1}=3 \cdot 4 \ldots \cdot n \cdot\left(a_{2}-a_{1}\right)=n!
$$

i.e.

$$
a_{n}=a_{n-1}+n!
$$

and therefore

$$
a_{n}=1!+2!+3!+\cdots+n!
$$

Now we have

$$
\begin{gathered}
a_{1}=1 \quad a_{2}=3 \quad a_{3}=9 \equiv-2 \\
a_{4} \equiv-2+2 \cdot 3 \cdot 4=0 \quad a_{5} \equiv 2 \cdot 3 \cdot 4 \cdot 5 \equiv-1 \quad a_{6} \equiv-1(-1) \cdot 6 \equiv-7 \equiv 4
\end{gathered}
$$

and in a similar way $a_{7} \equiv 6, a_{8} \equiv 0, a_{9} \equiv 1, a_{10} \equiv 0$. All congruences are modulo 11. As for $n \geq 11$ we have $n!\equiv 0$, from (1) it follows that then $a_{n} \equiv 0$, as well as for $n=4,8$.