## Task 1 - 200931

In the following schema, the letters are to be replaced by digits so that a correctly calculated division problem without a remainder is formed. Different letters can also be replaced by the same digits. As usual, a multi-digit number may not start with the digit 0.

Prove that there is exactly one such substitution that meets the requirements of the task! Determine this substitution!

$$
\begin{array}{ccccccccccccc}
a & b & c & 5 & 5 & : & 5 & d & e & = & f & 5 & g \\
h & i & 5 & & & & & & & & & & \\
\cline { 1 - 4 } j & k & m & n & & & & & & & & & \\
p & 5 & q & r & & & & & & & & & \\
\hline & s & t & u & v & & & & & & & & \\
& w & x & y & z & & & & & & & & \\
& & & & & & & &
\end{array}
$$
Due to $5 d e \geq 500$ and thus $2 \cdot 5 d e \geq 1000$, but $0<h i 5<1000$, it follows that $f=1$ and $h i 5=5 d e$, so $h=5$, $d=i$ and $e=5$.

Since $500<5 d 5<600$, it follows that $2500<5 \cdot 5 d 5<3000$, so $p=2$. Therefore, $5 \cdot 5 d 5<2600$, meaning $5 d 5<520$, so $d \in\{0,1\}$. Additionally, $5 \cdot 5 d 5$ ends in the units digit 5, so $r=5$ must also hold. This particularly results in the units digit of the difference $u=0$.

Due to the procedure in long division, $v=5$ must hold. To ensure no remainder, $w x y z=$ stuv must be true.

If all the information found so far is entered, one gets

| $a$ | $b$ | $c$ | 5 | 5 | $:$ | 5 | $d$ | 5 | $=$ | 1 | 5 | $g$ |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| 5 | $d$ | 5 |  |  |  |  |  |  |  |  |  |  |
| $j$ | $k$ | $m$ | 5 |  |  |  |  |  |  |  |  |  |
| 2 | 5 | $q$ | 5 |  |  |  |  |  |  |  |  |  |
|  | $s$ | $t$ | 0 | 5 |  |  |  |  |  |  |  |  |
|  | $s$ | $t$ | 0 | 5 |  |  |  |  |  |  |  |  |

If $d=0$, then $s t 05=g \cdot 505$ would have to hold. However, this contradicts $s \neq 0$, as the tens digit of $g \cdot 505$ is only 0 for the positive digit $g=1$.

Therefore, $d=1$ must be true. Then $s t 05=g \cdot 515$, which is exactly satisfied for the digit $g=7$, because then $7 \cdot 515=3605$. Also, $25 q 5=5 \cdot 515=2575$, so $q=7$ and thus $j k m 5=2575+360=2935$ and finally $a b c=515+293=808$, so the correctly calculated and uniquely determined division problem

| 8 | 0 | 8 | 5 | 5 | $:$ | 5 | 1 | 5 | $=$ | 1 | 5 | 7 |
| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |
| 5 | 1 | 5 |  |  |  |  |  |  |  |  |  |  |
| 2 | 9 | 3 | 5 |  |  |  |  |  |  |  |  |  |
| 2 | 5 | 7 | 5 |  |  |  |  |  |  |  |  |  |
|  | 3 | 6 | 0 | 5 |  |  |  |  |  |  |  |  |
|  | 3 | 6 | 0 | 5 |  |  |  |  |  |  |  |  |

is obtained, as one can easily verify.