In what number system is the following multiplication correct?

$166 \cdot 56=8590$.
Solution. Let the base of the number system be $a$. Since the number 9 appears in the result, $a > 9$. Furthermore, $6 \cdot 6 = 36$ and the product ends in 0, so the base must be a divisor of 36 that is greater than 9, i.e., $12, 18, 36$. Rewrite the multiplication in the decimal system:

$$
\left(a^{2}+6 a+6\right)(5 a+6)=8 a^{3}+5 a^{2}+9 a
$$

Performing the indicated operations and rearranging:

$$
f(a)=3 a^{3}-31 a^{2}-57 a-36=0
$$

For $a=12$, the value of (1) is:

$$
3 \cdot 12^{3}-31 \cdot 12^{2}-57 \cdot 12-36=5184-4464-684-36=0
$$

Thus, 12 is a solution to the problem.

We will show that there are no other solutions. For this, we only need to calculate the values of $f(18)$ and $f(36)$. Transform the function (1) into a product form.

$$
f(a)=(a-12)\left(3 a^{2}+5 a+3\right)
$$

Transform the quadratic expression, which represents a parabola, in the usual way to get:

$$
3 a^{2}+5 a+3=3\left[\left(a+\frac{5}{6}\right)^{2}+\frac{11}{36}\right]
$$

From this, we can see that the parabola lies above the $x$-axis, so the equation indeed has no other roots.