Which is the two-digit number that, when divided by the digit in the units place, gives a quotient of 9 and a remainder of 6?
I. Solution: Let the sought two-digit number be $10 x+y$, where, according to the requirements of the problem,

$$
x6 \text {. }
$$

According to the problem,

$$
10 x+y=9 y+6
$$

which means

$$
5 x-4 y=3
$$

Determining the integer solutions of this equation in the usual way,

$$
y=\frac{5 x-3}{4}=x+\frac{x-3}{4}=x+u
$$

where

$$
u=\frac{x-3}{4}
$$

which means

$$
x=4 u+3
$$

and thus

$$
y=x+u=5 u+3
$$

The inequalities under (1) are clearly only satisfied when $u=1$.

Therefore,

$$
x=7, \quad y=8
$$

and thus the sought number is 78.

István Cserteg (Budapest, VIII., Széchenyi g. III. o. t.)

II. Solution: Adding $y$ to both sides of equation (2)

$$
10 y=10 x+(2 y-6)
$$

it becomes clear that the expression in parentheses is divisible by 10. But since $6<y \leq 9$, the value of this expression can only be 10, that is,

$$
2 y-6=10
$$

from which

$$
y=8
$$

and thus the sought number is $9 y+6=78$.

István Boné (Budapest, XI., József A. g. II. o. t.)

III. Solution: Equation (2) can also be written as:

$$
x+y-6=9(y-x)
$$

Since $x$ and $y$ are digits, the left side is an integer between -6 and 12, so it can only be 0 or 9. The former case cannot hold, because then $x=y$ and the quotient would be 11, so

$$
\begin{gathered}
x+y-6=9 \\
y-x=1
\end{gathered}
$$

From this,

$$
x=7, \quad y=8
$$

Péter Lukács (Kecskemét, Katona J. g. IV. o. t.)