Solution

a) We are looking for a five-digit number $A B C D E$ such that

$$
A B C D E \times 9=E D C B A
$$

Adding $A B C D E$ to both sides of this equality, we get that $A B C D E \times 10 = E D C B A + A B C D E$. But the number $A B C D E \times 10$ can also be represented by the digits $A B C D E 0$. Thus, we have that $E D C B A + A B C D E = A B C D E 0$. To better visualize this, we show this equality as follows:

$$
\begin{array}{rccccc} 
& E & D & C & B & A \\
+ & A & B & C & D & E \\
\hline A & B & C & D & E & 0
\end{array}
$$

When adding two numbers with five digits and getting a result with six digits, the first digit of the result must be 1. Thus, we have that $A=1$:

$$
\begin{array}{rccccc} 
& E & D & C & B & 1 \\
+ & 1 & B & C & D & E \\
\hline 1 & B & C & D & E & 0
\end{array}
$$

Since $A=1$ and the last digit of the sum result is 0, we must necessarily have $E=9$:

$$
\begin{array}{rrrrrr} 
& & & & 1 \\
& 9 & D & C & B \\
+ & 1 & B & C & D & 9 \\
\hline 1 & B & C & D & 9 & 0
\end{array}
$$

which implies that:

$$
\begin{array}{rrrr} 
& & & 1 \\
& D & C & B \\
+ & B & C & D \\
\hline B & C & D & 9
\end{array}
$$

Now, we have two possibilities for the value of $B$. We must then have $B=1$ or $B=0$.

In the case where $B=1$, we have:

$$
\begin{aligned}
& 1 \\
& \begin{array}{lll}
D & C & 1
\end{array} \\
& \begin{array}{cccc}
+ & 1 & C & D \\
\hline 1 & C & D & 9
\end{array}
\end{aligned}
$$

Thus, we must have $D=7$ and then:

or more simply:

$$
\begin{array}{r}
7 C \\
+\quad 1 C \\
\hline 1 C 7
\end{array}
$$

which is impossible, since the sum of two numbers ending with the same digit must result in an even number.

Let's focus on the case $B=0$, from which we then get:

| $D$ | $C$ |
| :---: | :---: |
| + | $C$ |

Thus, we have $D=8$:

$$
\begin{array}{r} 
\\
8 C 0 \\
8 \quad \\
+\quad C 8 \\
\hline C 89
\end{array}
$$

which gives us $C=9$. Therefore, we conclude that $A B C D E=10989$.

b) The solution to this part is similar to the previous one. We are looking for a seven-digit number $A B C D E F G$ such that $A B C D E F G \times 9 = G F E D C B A$. Adding $A B C D E F G$ to both sides of this equality, we get that:

|  | $G$ | $F$ | $E$ | $D$ | $C$ | $B$ | $A$ |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| + | $A$ | $B$ | $C$ | $D$ | $E$ | $F$ | $G$ |
| $A$ | $B$ | $C$ | $D$ | $E$ | $F$ | $G$ | 0 |

Thus, we conclude that $A=1$ and $G=9$, obtaining:

|  |  |  |  |  | 1 |
| :---: | :---: | :---: | :---: | :---: | :---: |
|  | $F$ | $E$ | $D$ | $C$ | $B$ |
| + | $B$ | $C$ | $D$ | $E$ | $F$ |
| $B$ | $C$ | $D$ | $E$ | $F$ | 9 |

From this, we conclude that $B=1$ or $B=0$. If $B=1$ were correct, then to have the last digit of the result equal to 9, we would need $F=7$. Thus, the last sum shown would have a number starting with the digit 7 and another with the digit 1, and it could not result in a number with six digits. Therefore, we must abandon the possibility that $B=1$ and focus on the case $B=0$. In this case, we have $F=8$ and then the sum becomes:

$$
\begin{array}{r}
8 E D C \\
+\quad C D E \\
\hline C D E
\end{array}
$$

We now have two possibilities: $C=8$ or $C=9$. If $C=8$ were correct, then we would have $E=0$. In this case, the sum above would be:

| 80 D 8 |
| ---: |
| $+\quad 8 D 0$ |
| 8 D 08 |

which is impossible, since for $D+D$ to end with the digit 0, we would need $D=5$ and thus:

| 8058 |
| ---: |
| $+\quad 850$ |
| 8508 |

which is clearly wrong! Therefore, we must abandon the possibility that $C=8$ and focus on the case $C=9$. In this case, we have $E=9$, so:

![](https://cdn.mathpix.com/cropped/2024_05_01_463eeeabc25d438954b1g-28.jpg?height=238&width=357&top_left_y=1195&top_left_x=932)

which can only be satisfied if $D=9$.

Thus, the only possibility is $A B C D E F G=1099989$.
The number 1089 has an interesting property. When we multiply this number by 9, as shown below,

| 1 | 0 | 8 | 9 |
| :--- | :--- | :--- | :--- |
|  |  | $\times$ | 9 |
| 9 | 8 | 0 | 1 |

we get the number 9801, which is the number 1089 with its digits written from left to right!

a) Find a five-digit number $A B C D E$ such that its multiplication by 9 is equal to the number that has the digits of $A B C D E$ written from right to left, that is,

| $A$ | $B$ | $C$ | $D$ | $E$ |
| :---: | :---: | :---: | :---: | :---: |
|  |  |  | $\times$ | 9 |
| $E$ | $D$ | $C$ | $B$ | $A$ |

b) Find all the seven-digit numbers whose multiplication by 9, as before, reverses the position of their digits.