## Task 28/65

A two-digit number is to be found, where the product of the two digits is equal to the difference between five times the square of the last digit and the cross sum increased by 10.
The sought number is $z=10 x-y$ with $0 \leq x ; y \leq 9$, both integers. From the conditions of the problem, the equation follows:

$$
x y=5 y^{2}-x-y-10 \quad \text { or } \quad y^{2}=\frac{x y+x+y+10}{5}
$$

Since $x y+x+y$ is certainly less than 100 due to the conditions for $x$ and $y$, it follows that $y^{2}<22$ and thus $y \leq 4$. Among the possible values $y=1 ; 2 ; 3 ; 4$, only the value $y=3$ with $x=8$ satisfies the equation. Therefore, the only solution is the pair with $x=8, y=3$, and thus $z=83$.

![](https://cdn.mathpix.com/cropped/2024_06_06_1d2be4837fb839c2b31dg-120.jpg?height=1274&width=1608&top_left_y=1399&top_left_x=224)