Task B-1.5. If we add the sum of the digits of a two-digit number to the square of that sum, we will get the two-digit number again. Determine all such two-digit numbers.
## Solution.

Let the two-digit number be $\overline{x y}=10 x+y$.

Then we have

$$
x+y+(x+y)^{2}=10 x+y, \quad x \neq 0
$$

From this, $(x+y)^{2}=9 x$.

Thus, the product of the first digit and the number 9 is the square of the sum of the digits.

We will easily list all products of the number 9 with $1,2,3,4,5,6,7,8,9$ and check which of them are squares of natural numbers. Squares are obtained for $x=1, x=4$ and $x=9$.

$9 \cdot 1=9, 9 \cdot 4=36, 9 \cdot 9=81$.

(2 points)

For $x=1$, from $(x+y)^{2}=9 x$ we get $y=2$, and $\overline{x y}=12$.

Similarly, for $x=4$, $\overline{x y}=42$ and for $x=9$, $\overline{x y}=90$.

(2 points)

The sought two-digit numbers are: 12, 42, and 90.

Note.

If someone guessed all three numbers without the process, give 3 points. If someone listed all two-digit numbers and thus obtained the solution, give 6 points.