1. Eva, Igor, Marko, and Maruša each wrote a natural number on a piece of paper. If they erased the last digit of Eva's number, they would get Igor's number. If they erased the last digit of Igor's number, they would get Marko's number. If they erased the last digit of Marko's number, they would get Maruša's number. The sum of all four written numbers was 3838. What numbers did Eva, Igor, Marko, and Maruša write?
I/1. Evino number must be at least four-digit, so that we can consecutively remove the last digit three times and still get a natural number. However, Evino number cannot be more than four-digit, since the sum of all four numbers is four-digit. Let's denote Evino number as $\overline{a b c d}$, where $a, b, c$, and $d$ are digits. Then Igor's number is $\overline{a b c}$, Mark's number is $\overline{a b}$, and Maruša's number is $\bar{a}$. The sum of all four numbers is

$$
1000 a+100(a+b)+10(a+b+c)+(a+b+c+d)=3838
$$

Since $a, b, c$, and $d$ are digits, $a+b+c+d \leq 36$. Since the units of the number 3838 are 8, $a+b+c+d \leq 28$. Therefore, at most 2 is carried over to the tens in the calculation $\mathrm{k}$. Since $a+b+c+2 \leq 29$, at most 2 is carried over to the hundreds. Since $a+b+2 \leq 20$ and the hundreds of the number 3838 are 3, at most 1 is carried over to the thousands. From this, it follows that $a$ is equal to 2 or 3. The possibility $a=2$ is ruled out, because in this case $10 \leq a+b=2+b \leq 11$, so at least 7 would have to be carried over to the hundreds, which is not possible. Therefore, $a=3$.

Since at most 2 is carried over to the hundreds, $b$ is equal to 3, 4, or 5. The possibility $b=5$ is ruled out, because in this case $8 \leq 8+c=a+b+c \leq 9$, so at least 4 would have to be carried over to the tens, which is not possible. The possibility $b=3$ is also ruled out. In this case, 2 would have to be carried over to the hundreds. But since at most 2 is carried over to the tens, $a+b+c+2=8+c<20$ and therefore at most 1 would be carried over to the hundreds. Therefore, $b=4$.

Since at most 2 is carried over to the tens, $c$ is equal to 4, 5, or 6. The possibility $c=6$ is ruled out, because in this case $13+c=a+b+c+d \leq 9$, which is not true. If $c=4$, 2 would have to be carried over to the tens. In this case, we would have $a+b+c+d=11+d$, so $d$ would have to be 9. But then the units of the sum of the numbers would be 0, which is a contradiction. Therefore, $c=5$ and thus $d=6$.

Eva, Igor, Marko, and Maruša wrote the numbers $3456, 345, 34$, and 3.
Justification of how many digits the numbers have ..... 1 point
Justification that $a=3$ ..... 2 points
Justification that $b=4$ ..... 2 points
Justification that $\mathrm{c}=5$ and $\mathrm{d}=6$ ..... 1 point
Written solution (all four numbers) ..... 1 point