## Task 1 - 130931

As can be seen from examples, there are pairs $(x, y)$ where $x$ and $y$ are each a two-digit natural number with the following property:

If the digits of this number are swapped and 9 is added to the resulting number, the other number of the pair is obtained. (An example of such a pair is (25; 61), since $52+9=61$ and $16+9=25$.)

Hint: If swapping the digits results in a number sequence starting with 0 (e.g., from 30 to "03"), the (single-digit) number that remains after removing the zero should be used for further operations (in our example, "3").

We call the numbers $x, y$ of such a pair $(x ; y)$ associated with each other.

a) List all two-digit numbers that can appear as elements of such pairs!

b) Determine all two-digit numbers that are associated with themselves!
a) If $10 a+b$ and $10 c+d$ are the two numbers of such a pair, with $a, b, c, d \in\{0, 1, 2, \ldots, 9\}$ and $a \neq 0, c \neq 0$ (otherwise, they would not be two-digit numbers), then it holds that $10 c+d=10 b+a+9$. Since $1 \leq a<10$, it follows that $10 \leq a+9<19$, so the tens digit of $10 b+a+9$ is $b+1$ and the units digit is $a+9-10=a-1$. We thus obtain $c=b+1$ and $d=a-1$. Since $a \geq 1$, $d$ is always a digit, so no additional conditions arise here. However, because $c=b+1$, $b \leq 8$ must hold, as for $b=9$ one would get $c=10$, which is no longer a digit.

In fact, these conditions are not only necessary but also sufficient: Applying the procedure to $10 c+d$ yields the number $10 d+c+9=10(a-1)+(b+1)+9=10 a-10+b+9+1=10 a+b$; thus, the original number.

Therefore, exactly those two-digit numbers that do not have 9 as their units digit can be elements of such a pair.

b) With the above notations, it must then hold that $a=c=b+1$ and $b=d=a-1$, so this is only the case for the numbers 21, 32, 43, 54, 65, 76, 87, and 98. The verification confirms that these numbers are indeed each assigned to themselves.