Problem 882

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 $n=4,8$.