## Task 3 - 290523

A natural number $z$ is sought that satisfies the following conditions:

(1) The digit in the tens place of $z$ is 0.

(2) If one forms a new number $z^{\prime}$ by removing the digit 0 in the tens place from $z$ and then calculates the sum $z+z^{\prime}$, the result is 5174.

Show that there can only be one number that satisfies these conditions, and give this number! Also, verify that the number $z$ you provide satisfies the conditions!
If a number $z$ satisfies the conditions, it must be a four-digit number; for if it had more digits, then $z+z^{\prime}$ would certainly have more digits; but if it had 3 or fewer digits, then $z+z^{\prime}$ could be at most a four-digit number starting with the digit 1.

Now, if $a, b, c, d$ are the digits of $z$, it follows from (1) that $c=0$, and due to (2), the cryptogram

![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-0125.jpg?height=128&width=326&top_left_y=2226&top_left_x=868)

is satisfied. At the units place, it is evident that $2 d=4$ or $2 d=14$.

In the case $2 d=4$, i.e., $d=2$, it follows for the tens place that $b=7$, and at the hundreds place, due to $7+a=11$, only $a=4$ can stand.

In the case $2 d=14$, i.e., $d=7$, it follows for the tens place that $b=6$, and for the hundreds place, due to $6+a=11$, $a=5$ would follow, and because of the carry, a 6 would result in the thousands place of the sum, not a 5. Therefore, the case $2 d=14$ is excluded.

Thus, the conditions can only be satisfied by the number $z=4702$. The verification

$$
\begin{array}{r}
47102 \\
+\quad 472 \\
\hline 5174
\end{array}
$$

shows that this number satisfies the conditions.