Italian Mathematical Olympiad - February Round

Solution:

The answer is $(\mathbf{C})$. Let us consider the equations
$$\left\{ \begin{array}{l} a_{n}=n\left(a_{1}+a_{2}+\cdots+a_{n-1}\right) \\ a_{n+1}=(n+1)\left(a_{1}+a_{2}+\cdots+a_{n-1}+a_{n}\right) \end{array} \right.$$
valid for every $n \geq 2$. Let us now rewrite the second one in the form
$$\begin{aligned} a_{n+1} & =n\left(a_{1}+a_{2}+\cdots+a_{n-1}\right)+\left(a_{1}+a_{2}+\cdots+a_{n-1}\right)+(n+1) a_{n} \\ & =a_{n}+\frac{a_{n}}{n}+(n+1) a_{n} \\ & =\frac{n^{2}+2 n+1}{n} a_{n} \\ & =\frac{(n+1)^{2}}{n} a_{n} \end{aligned}$$
where - using the first equation - we have replaced $a_{1}+a_{2}+\cdots+a_{n-1}$ with $\frac{a_{n}}{n}$ wherever possible. We then obtain
$$a_{n}=\frac{n^{2}}{n-1} a_{n-1}=\frac{n^{2}}{n-1} \frac{(n-1)^{2}}{n-2} a_{n-2}=\cdots=\frac{n^{2}}{n-1} \frac{(n-1)^{2}}{n-2} \cdots \frac{3^{2}}{2} a_{2} .$$
Simplifying the denominator of each fraction with the numerator of the next one, and observing that $a_{2}=2$, we obtain
$$a_{n}=n^{2} \cdot(n-1) \cdot(n-2) \cdot \ldots \cdot 4 \cdot \frac{3}{2} a_{2}=n \cdot \frac{n !}{2}$$
for every $n \geq 2$. At this point, we observe that $2022=2 \cdot 3 \cdot 337$ divides $n \cdot \frac{n !}{2}$ if and only if this number is divisible separately by 2, 3 and 337. On one hand, since $n \cdot \frac{n !}{2}$ is a product of numbers less than or equal to $n$ and 337 is prime, such a product can be divisible by 337 only if $n \geq 337$. On the other hand, if $n=337$ we have $a_{n}=337 \cdot \frac{1}{2} \cdot 337!$, and $\frac{1}{2} \cdot 337! = 3 \cdot 4 \cdot \ldots \cdot 337$ is certainly a multiple of 6, so that $2 \cdot 3 \cdot 337$ divides $a_{337}$, and therefore 337 is the minimum sought.