13.308. What two-digit number is less than the sum of the squares of its digits by 11 and greater than their doubled product by 5?
Solution.

Let $10 x+y$ be the desired number. According to the condition, we have the system $\left\{\begin{array}{l}x^{2}+y^{2}=10 x+y+11 \\ 2 x y=10 x+y-5\end{array}\right.$

Subtracting the second equation from the first, we find $(x-y)^{2}=16$ and obtain two systems:

$$
\left\{\begin{array} { l } 
{ x - y = 4 } \\
{ 2 x y = 1 0 x + y - 5 }
\end{array} \text { and } \left\{\begin{array}{l}
x-y=-4 \\
2 x y=10 x+y-5
\end{array}\right.\right.
$$

The solutions of which are $x=9, y=5$ and $x=1, y=5$.

Answer: 95 or 15.