Standa and Jana received two three-digit numbers. Standa added a decimal point after the first digit in the first number, and after the second digit in the second number, then added the resulting decimal numbers and got the result 50.13. Jana added a decimal point after the second digit in the first number, and after the first digit in the second number, then added the resulting decimal numbers and got the result 34.02.

Determine the sum of the original three-digit numbers.

(K. Pazourek)
Stand's, or Janin's calculation can be clearly written as follows:

![](https://cdn.mathpix.com/cropped/2024_04_17_cbd86f0d05e49142d45cg-2.jpg?height=175&width=470&top_left_y=418&top_left_x=793)

By comparing the values in the hundredths place in the first, or second calculation, we get $c=3$, or $f=2$. The previous calculations can be expressed as follows:

$$
\begin{gathered}
a, b 3 \\
d e, 2
\end{gathered} \quad \begin{gathered}
a b, 3 \\
d, e 2
\end{gathered}
$$

By comparing the values in the tenths place in the first, or second calculation, we get $b=9$, or $e=7$. The previous calculations can be expressed as follows, keeping in mind that in both cases there is a carry-over of $\mathrm{k}$ to the tens place:

$$
\begin{array}{r}
a, 93 \\
d 7,2
\end{array} \quad \begin{array}{r}
a 9,3 \\
d, 72 \\
\hline 50,13
\end{array} \quad \begin{array}{r}
34,02
\end{array}
$$

By comparing the values in the units place in the first, or second calculation, we get $a=2$, or $d=4$. After substitution, we find that the match holds even in the tens place:

$$
\begin{array}{cc}
2,93 \\
47,2
\end{array} \quad \begin{array}{r}
29,3 \\
4,72 \\
\hline 50,13
\end{array} \quad \begin{array}{r}
34,02
\end{array}
$$

The original three-digit numbers were 293 and 472. The sought sum is

$$
\begin{array}{lll}
2 & 9 & 3 \\
4 & 7 & 2 \\
\hline 7 & 6 & 5
\end{array}
$$

Evaluation. 1 point for a suitable problem statement; 3 points for the steps in determining the unknown digits; 2 points for the final calculation and result.