The sought three-digit number is $100x + 10y + z$. According to the problem,

$$
100x + 10y + z + x + y + z = 100a + 10a + a
$$

or

$$
100x + 10y + (x + y + 2z) = 111a, \ldots
$$

where $x, y, z, a$ are positive integers less than 10; $y$ and $z$ can also be zero. (Neither $a$ nor $x$ can be 0!)

From the very beginning, two cases are possible: in both equations, the number of hundreds is the same, i.e., $x = a$, or the number of tens resulting from the addition of the units and tens on the left side adds up to - at most - one hundred, in which case $x + 1 = a$, or $x = a - 1$.

I. In the case of $x = a$, equation 2) becomes:

$$
11y + 2z = 10a \quad \text{or} \quad z = 5a - \frac{11y}{2} \ldots
$$

We see, therefore, that $y$ can only be $0, 2, 4, 6, 8$.

For $y = 0$, $z = 5a$; only $a = 1$ is possible, and thus $x = 1, z = 5$. (105!)

For $y = 2$, $z = 5a - 11$; since $0 \leq 5a - 11 \leq 9$, only $a = 3$ and 4 are possible, and thus $z = 4$, or 9.

The corresponding numbers are 324 and 429.

For $y = 4$, $0 \leq z = 5a - 22 \leq 9$, so $a = 5$, or 6. Thus, $x = 5$, or 6, and $z = 3$, or 8.

The corresponding numbers are 543 and 648.

For $y = 6$, $0 \leq z = 5a - 33 \leq 9$; hence $a = 7$, or 8. Thus, $x = 7$, or 8, and $z = 2$, or 7.

The corresponding numbers are 762 and 867.

For $y = 8$, $0 \leq z = 5a - 44 \leq 9$; hence only $a = 9, x = 9, z = 1$.

The corresponding number is 981.

II. In the case of $x = a - 1$, $a \geq 2$. Then from equation 2), we get:

$$
11y + 2z = 10a + 101, \quad \text{or} \quad z = 5a + \frac{101 - 11y}{2} \ldots
$$

Since $a$ is at least 2 and $y$ is at most 9, clearly $z > 10$.

Therefore, there is no such solution, and the numbers that fit the problem are:

105, 324, 429, 543, 648, 762, 867, 981.

György Holló (Áll. Dobó István g. VI. o. Eger)
Which three-digit numbers in the decimal system, when added to the sum of their digits, result in a number of the form $\overline{a a a}$?