Let's determine the base $x$ of the number system if the following equation holds:

$$
2016_{x}=x^{3}+2 x+342
$$
I. solution. $x$ is a positive integer greater than 6, because the number written in base $x$ contains the digit 6. Since $2016_{x}=2 x^{3}+x+6$, we can write the following equation:

$$
\begin{aligned}
2 x^{3}+x+6 & =x^{3}+2 x+342 \\
x^{3}-x-336 & =0
\end{aligned}
$$

Let's examine the solutions of the equation in the set of natural numbers; these can be chosen from the divisors of the constant term. The positive divisors of -336 in sequence: $1,2,3,4,6,7, \ldots$. It is enough to examine from 7, because $x>6$. The $x=7$ satisfies the equation. Therefore, $(x-7)$ can be factored out from the expression on the left side of the equation:

$$
x^{3}-x-336=(x-7)\left(x^{2}+7 x+48\right)=0
$$

Since a product is zero if and only if one of its factors is zero, we still need to check when the quadratic expression is zero. But the discriminant of the quadratic equation $x^{2}+7 x+48=0$ is $D=7^{2}-4 \cdot 48<0$, so it has no real roots.

Thus, the only solution is $x=7$.

Rearranging the equation $x^{3}-x-336=0$:

$$
336=x^{3}-x=x\left(x^{2}-1\right)=x(x+1)(x-1)=(x-1) x(x+1)
$$

Zoárd Csóka (Győri Műszaki Szakképzési Centrum Jedlik Ányos Gépipari és Informatikai Szakgimn., Szki. és Koll., 10th grade)

Using this, the solution could also be reached in another way. We show some possible paths.

II. solution. Since $x \in \mathbb{N}$ (or $x \geq 7$ due to the 6 in $2016_{x}$), we are essentially looking for three consecutive positive integers whose product is 336. Since $\forall x_{1}<x_{2}\left(x_{1}, x_{2} \in \mathbb{N}\right)$

$$
\left(x_{1}-1\right) x_{1}\left(x_{1}+1\right)<\left(x_{2}-1\right) x_{2}\left(x_{2}+1\right)
$$

because the factors in the two products are always larger in pairs, the equation has only one real root. In the smallest possible case of $x=7$, the equality indeed holds $(6 \cdot 7 \cdot 8=336)$, so this is the only solution.

Dávid Apagyi (Kecskeméti Katona J. Gimn., 9th grade)

III. solution. Since $x$ is a positive integer, $x(x-1)(x+1)$ is the product of three consecutive positive integers.

The prime factorization of 336: $2^{4} \cdot 3 \cdot 7$. From this, it follows that $6 \cdot 7 \cdot 8=336$, so $x=7$ and no other solution is possible.

Gergő Demeter (Szegedi Radnóti M. Kísérleti Gimn., 11th grade)

IV. solution. From the last equation, we get that the product of three consecutive positive integers is 336. Since the number system allows the digit 6, $x \geq 7$.

$$
\sqrt[3]{(x-1) x(x+1)}=\sqrt[3]{336} \approx 6.95
$$

The geometric mean of the three numbers is $6.95 \ldots$, and the geometric mean cannot be smaller than all three numbers, so the smallest number is at most 6, thus $x \leq 7$. Therefore, $x$ can only be 7. To check if $x=7$ is correct, we can easily calculate: $6 \cdot 7 \cdot 8=336$, so this is indeed a valid solution.

Dániel Dobák (Budapest V. Ker. Eötvös J. Gimn., 10th grade)