Vojta wanted to add several three-digit natural numbers on a calculator. On the first attempt, he got the result 2224. To check, he added the numbers again and got 2198. He calculated once more, and this time the sum was 2204. It turned out that the last three-digit number was cursed - each time Vojta did not press one of its digits hard enough, and the calculator thus entered a two-digit number instead of the three-digit one. There were no other errors in the addition. What is the correct sum of Vojta's numbers?

(L. Šimůnek)
A cursed number will be denoted as $\overline{A B C}$. However, Vojta added two-digit numbers $\overline{A B}, \overline{A C}, \overline{B C}$, which we will generally denote as ??. We will denote as **** the sum of the numbers that Vojta could add without making a mistake. The schematic addition is expressed as follows:

![](https://cdn.mathpix.com/cropped/2024_04_17_27d0b340b16adf4cc1afg-6.jpg?height=156&width=767&top_left_y=379&top_left_x=650)

The first and third sums have the same digit in the units place, so in these cases, the number ?? should be $\overline{A C}$ and $\overline{B C}$. The remaining number $\overline{A B}$ thus belongs to the sum 2198. The sum 2224 is 26 greater than 2198, and therefore it could not have been formed by adding the number **** and the number $\overline{A C}$, which differs from the number $\overline{A B}$ only in the units digit. The numbers $\overline{A B}, \overline{A C}, \overline{B C}$ are thus uniquely assigned to the sums:

![](https://cdn.mathpix.com/cropped/2024_04_17_27d0b340b16adf4cc1afg-6.jpg?height=158&width=759&top_left_y=869&top_left_x=657)

According to the first and third sums, $B = A + 2$. According to the second and third sums, $C = B + 6$, and by substituting the previous relation, we get: $C = A + 2 + 6 = A + 8$. The letter $A$ is a non-zero single-digit number, and for $C$ to also be a single-digit number, $A$ can only be 1. Then $B = 3$ and $C = 9$. The cursed number was thus 139. The sum **** can be calculated, for example, as follows: $2224 - 39 = 2185$. Vojta should have correctly obtained the sum $2185 + 139 = 2324$.

Note. To solve the problem, it is not necessary to determine the digits $B$ and $C$. The conclusion can be reached without knowing these digits, i.e., without knowing the cursed number, as follows: $2224 + \overline{A 00} = 2224 + 100 = 2324$.

Evaluation. 2 points for correctly assigning the positions of the missing digits to the individual sums; 3 points for calculating the missing digits (in a certain type of solution, it may be sufficient to know only the digit $A$, see note); 1 point for the correct final sum.