206. Find all two-digit numbers, each of which is 13 more than the sum of the squares of its digits.
$\triangle$ Let the desired number be denoted as $\overline{x y}$. We obtain:

$$
10 x + y = x^2 + y^2 + 13, \quad x^2 - 10 x + (y^2 - y + 13) = 0
$$

We will consider the last equation as a quadratic equation in terms of $x$. Its discriminant must be non-negative:

$$
\frac{1}{4} D = 25 - y^2 + y - 13 \geq 0, \quad -y^2 + y + 12 \geq 0, \quad y^2 - y - 12 \leq 0, \quad -3 \leq y \leq 4
$$

Since $y$ is a digit, we find more precise bounds for $y$:

$$
0 \leq y \leq 4
$$

Now we will check the cases $y=0,1,2,3,4$.

Answer: 54.