## Task 1 - 291211

Determine the number of all natural numbers $z$ with the following properties:

(1) The decimal representation of $z$ consists of five pairwise distinct digits.

(2) The first and the last digit in it are different from 0.

(3) If $z^{\prime}$ is the number whose digit representation is obtained from that of $z$ by reversing the order, then the digit representation of the number $z+z^{\prime}$ consists of all identical digits.
The digits of $z$ are $a, b, c, d, e$. The task is to determine the number of solutions to a problem that, in the style of a cryptogram,

![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-2259.jpg?height=137&width=422&top_left_y=914&top_left_x=451)

![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-2259.jpg?height=140&width=668&top_left_y=912&top_left_x=934)

is stated. For each such solution, it follows that:

Since in the units place $a+e=f$ or $a+e=10+f \quad$ (4) is required, there can be no carry from the thousands place to the ten-thousands place, i.e., $b+d<10$, and thus, according to (4), $a+e=11$. Due to the carry that occurs in the units place and $f=1$, $b+d=0$ or $b+d=10$, and thus, according to (5), $b+d=0$ must hold. This would result in the contradiction $b=d=0$ against (1). Therefore, the case (**) is excluded.

In (*), it follows that $a+e<10 \quad$ (6) and thus, for (*) to be satisfied, without any carry at any place,

$$
0<f=a+e=b+d=2 c<10
$$

must hold. For the representation of the even numbers $f$ between 0 and 10 as the sum of two distinct natural numbers $a, e$ that are different from each other and from 0, there are exactly the possibilities listed in the table:

| $f$ | Representation | $c$ |
| :--- | :--- | :--- |
| 2 | - |  |
| 4 | $1+3$ | 2 |
| 6 | $1+5,2+4$ | 3 |
| 8 | $1+7,2+6,3+5$ | 4 |

For the representation as the sum of two distinct natural numbers $b, d$, where 0 can also be included, the possibility $f=0+f$ is added each time. The number $c$ corresponding to each $f$ is always different from the possibilities given for $a, e$ and $b, d$.

In the case $f=4$, (*) is satisfied exactly with the sets (unordered pairs) $\{a, e\}=\{1,3\}$ and $\{b, d\}=\{0,4\}$. In each of these sets, there are exactly 2 possibilities for the order; thus, (*) has exactly 4 solutions for $f=4$.

In the case $f=6$, for $\{a, e\}$, one of the sets $\{1,5\},\{2,4\}$ (4 possibilities) and for $b, d$ either the other of these sets or 0, 6 (4 possibilities) must be taken; thus, (*) has exactly 16 solutions for $f=6$.

In the case $f=8$, it follows similarly that there are 6 possibilities for $\{a, e\}$ and 6 possibilities for $\{b, d\}$, so (*) has exactly 36 solutions for $f=8$.

The total number of solutions sought is therefore 56.

Adapted from $[5]$