## Task 3 - 170513

Fritz wants to write down a subtraction problem where the difference between two natural numbers is to be formed.

The result should be a three-digit number whose three digits are all the same. The minuend should be a number that ends in zero. If this zero is removed, the subtrahend should result.

Give all subtraction problems for which this is true!
Let $m$ be the minuend ending in zero, and $s$ be the subtrahend. If we remove the zero from $m$, we get $s$, meaning the minuend is ten times the subtrahend: $m=10 s$. The difference $m-s$ is thus $10 s-s=9 s$.

The difference is a three-digit number where all digits are the same. Let $n$ be this digit $(n=1,2, \ldots, 9)$, i.e.,

$$
9 s=100 n+10 n+n=111 n \quad \Rightarrow \quad 3 s=37 n
$$

Since 37 is not divisible, 3 must be divisible. Trying the possible values $n=3,6,9$, we get

$n=3 \Rightarrow s=37, m=370 \Rightarrow 370-37=333$

$n=6 \Rightarrow s=74, m=740 \Rightarrow 740-74=666$

$n=9 \Rightarrow s=111, m=1110 \Rightarrow 1110-111=999$.

Thus, there are three possible subtraction problems.

Official solution: If there is such a subtraction problem, then in it the minuend is ten times the subtrahend. Therefore, the difference is nine times the subtrahend.

The only three-digit numbers that are divisible by 9 and consist of identical digits are 333, 666, and 999.

Since $333 \div 9 = 37, 666 \div 9 = 74, 999 \div 9 = 111$, the subtrahend can only be one of the numbers 37, 74, 111. Therefore, the following subtraction problems can meet the conditions of the problem:

$$
\begin{array}{rrr}
370 \\
-37 \\
\hline 333
\end{array} \quad \begin{array}{r}
740 \\
-\quad 74 \\
\hline 666
\end{array} \quad \begin{array}{r}
1110 \\
-111 \\
\hline 999
\end{array}
$$

For these, all the requirements of the problem are met. Therefore, these three are the sought-after subtraction problems.