\section*{Task 3 - 221013}

Ten Little Digits

Wilhelm Busch set an example through the diligent teacher Lämpel. How one should toil, he once had him say: "Not only in writing, reading does a sensible being practice, but also in arithmetic, a person should make an effort." Reading this backwards, it is surely also of great value: "Not only in arithmetic should a person make an effort, but a sensible being should also sometimes read." Therefore, do not hesitate, read this poem first!

First of all, it should be said immediately that numbers are being asked for. Among these are two three-digit addends. We assume the sum

(since this is possible in many cases) has one more digit.

Now it is surely not very difficult to count that ten digits are good for this case:

The digits are from 0 to 9, which delight us for this purpose! And each digit appears in this calculation once, and exactly (as one says)! So that no one asks later: The 0 would not fit well at the front, so it is not allowed there. There is exactly one carry, not two or three!

(A carry - this should be explained so that everyone can learn - is a case that occurs when someone adds numbers and their sum, as it turns out, exceeds 9 in size.)

Now, dear daughter, dear son, what results can one expect from this addition?

You may now start solving. As a solution, at least one should strive to provide all possible final sums! Here, one should vigorously prove that it is only these sums that can be, if the rules are followed.

(The multiple possibilities of the addends should not worry us, they are \(\mathrm{n}\) o t asked for here.)

Now get to work and do not be discouraged! For not only in reading does a sensible being practice ...
}

Solution:

In the result with four digits, every solver will equally realize that only the 1 can be at the front: It comes from the carry into the "hundreds" column

(If it were different, one would be surprised.) The 0 cannot be present in the addends here at all, as no other digit can be taken twice; as has just been revealed, the carry is already "given away." Also, in the units and tens, 0 cannot stand in the result, otherwise there would be a second carry, which, as we know, no one can.

Now the solution is half won, it starts with 1 and 0.

![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-1456.jpg?height=123&width=304&top_left_y=2537&top_left_x=885)

Now one thinks about the addends, what is present in the hundreds?

At first, one might have thought of the digits 2 and 8, but 3 and 7 also work; and 4 and 6 are still left.

Therefore, one makes a kind of table "on the fly" from these:

\begin{tabular}{llll}
\begin{tabular}{l} 
Used \\
Digits
\end{tabular} & \begin{tabular}{l} 
Remaining \\
Digits
\end{tabular} & \begin{tabular}{l} 
Only usable \\
Sum formations
\end{tabular} \\
\hline \(0,1,2,8\) & \(3,4,5,6,7,9\) & - \\
\(0,1,3,7\) & \(2,4,5,6,8,9\) & \(2+6=8\) & \(4+5=9\) \\
\(0,1,4,6\) & \(2,3,5,7,8,9\) & \(2+7=9\) & \(3+5=8\)
\end{tabular}

Finally, one must try to combine the rest in such a way that no carry is generated when one starts adding.

If one thinks about it for a while, one sees in the first row that it does not work, no matter what one tries.
The second and third rows then each offer a solution: First, 1, 0, 9, and 8.

And, if one does it differently, one can also be delighted with \(1,0,8\) and 9.

These two sums are the only possible ones. Every child can see that it can only be this way if one follows the rules.