Let's determine those two-digit numbers, which when divided by the sum of their digits, yield a quotient of 7 and a remainder of 6.
Let the sought two-digit number be $\overline{x y}=10 x+y$. Then, according to the problem's conditions,

$$
10 x+y=7 \cdot(x+y)+6
$$

and here

$$
66, \quad \text{i.e.,} \ y>\frac{4}{3}
$$

Furthermore, based on (2) and (3),

$$
2 y+2 \leqq 9, \quad \text{i.e.,} \ y \leqq \frac{7}{2}
$$

Since $y$ is an integer, only the values 2 and 3 satisfy both constraints. For $y=2$, $x=6$, and for $y=3$, $x=8$. Therefore, the sought two-digit numbers are 62 and 83. These numbers also satisfy the problem's requirements.

![](https://cdn.mathpix.com/cropped/2024_05_02_006886fb8e6f3ee56d0cg-1.jpg?height=631&width=702&top_left_y=1186&top_left_x=706)

Remarks. 1. If the integers $x, y$ are considered as the coordinates of a point $P$ in a Cartesian coordinate system, their values can also be determined graphically. The possible positions of the point $P$ are determined by the inequalities (1) and (2), as well as the equation (1). The inequalities (2) mean that the point $P$ can only be to the right of the line $x=1$, to the left of the line $x=9$ (Figure 2), above the line $y=0$, and below the line $y=9$, or on the boundary lines. According to (1), the point $P$ can only be in the direction of larger ordinates from the line $x+y=6$. The possible positions of the point $P$ are indicated by hatching. Finally, (3) means that the point $P$ can only be on the integer-coordinate points of the line

$$
y=\frac{x}{2}-1
$$

From the diagram, it can be seen that there are only two points that satisfy all conditions, $P_{1}(6,2)$ and $P_{2}(8,3)$, so the sought numbers are 62 and 83.

2. Many contestants incorrectly considered the numbers 20 and 41 as solutions because $20=2 \cdot 7+6$ and $41=5 \cdot 7+6$, and did not realize that the remainder of a division cannot be greater than the divisor.