We add the sum of the digits of a two-digit number to the number itself, and then add the sum of the digits of the resulting number to it again. This way, we arrive at a two-digit number whose digits are the digits of the original number, but in reverse order. Which number is this?

 

We add the sum of the digits of a two-digit number to the number itself, and then add the sum of the digits of the resulting number to it again. This way, we arrive at a two-digit number whose digits are the digits of the original number, but in reverse order. Which number is this?
I. solution. Let's denote the number we are looking for as: $N=\overline{x y}=10 x+y$, here $x \geq 1$, because the number is two-digit. Adding the sum of the digits $x+y$ to $N$ gives:

$$
N_{1}=10 x+y+(x+y)=11 x+2 y
$$

The last digit of the sum $M=x+2 y$ will be the last digit of $N_{1}$, and the first digit of $N_{1}$ will be $x$ plus the tens digit of $M$. Let's denote this digit by $k$, for which $0 \leq k \leq 2$, because $M \leq 3 \cdot 9=27$.

The last digit of $M$ and thus of $N_{1}$ is $M-10 k=x+2 y-10 k$, the first digit of $N_{1}$ is $x+k$, so the sum of the digits of $N_{1}$ is $x+k+x+2 y-10 k=2(x+y)-9 k$. Adding this to $N_{1}$, the resulting number $N_{2}$ should be the number $10 y+x$ formed by swapping the digits:

$$
N_{2}=11 x+2 y+2(x+y)-9 k=13 x+4 y-9 k=10 y+x
$$

From this,

$$
6 y+9 k=12 x, \quad 2 y+3 k=4 x
$$

Here, $3 k$ and thus $k$ itself can only be even, so the value of $k$ is 0 or 2.

If $k=0$, then $y=2 x$, the sum of the digits of $N$ is $x+2 y=5 x$ and is a single digit, so $x=1, y=2$, and $N=12$ satisfies the conditions of the problem, because

$$
N_{1}=12+1+2=15, \quad N_{2}=15+1+5=21
$$

If $k=2, y=2 x-3$, the sum of the digits of $N$ is

$$
20 \leq x+2 y \leq 3 \cdot 9=27, \quad 20 \leq 5 x-6 \leq 27, \quad 26 \leq 5 x \leq 33
$$

According to this, only $x=6$ is possible, so $y=9$, and $N^{\prime}=69$ is also a solution to the problem, because for it $N_{1}^{\prime}=69+6+9=84$, $N_{2}^{\prime}=84+8+4=96$.

II. solution. Let's denote the digits of the number formed by adding the digits of the sought $10 x+y$ number by $a, b$:

$$
10 x+y+x+y=11 x+2 y=10 a+b
$$

According to the condition, adding its digits to this should result in the number $10 y+x$:

$$
10 a+b+a+b=11 a+2 b=10 y+x
$$

From the two relationships, we eliminate $x$:

$$
\begin{aligned}
11(10 y+x)-(11 x+2 y) & =108 y \\
& =11(11 a+2 b)-(10 a+b)=111 a+21 b
\end{aligned}
$$

From this,

$$
36 y=37 a+7 b, \quad y=a+\frac{a+7 b}{36}
$$

Here, the numerator of the last fraction must be divisible by 36 because $y$ is an integer, and $a$ is at most 8 because $y$ is larger and a digit. Thus, $a+7 b \leq 8+7 \cdot 9=71$, so $a+7 b=36$ must hold. From this,

$$
b=5-\frac{a-1}{7}
$$

This can only be an integer if $a-1$ is divisible by 7, which only follows for $a=1$ and $a=8$ among the digits; the corresponding values of $b$ are 5 and 4. The number formed by swapping the digits of the sought number is then $11 a+2 b=21$, or 96, so the sought number can be 12 and 69, and both satisfy the conditions of the problem.

III. solution. We show that the sought number is divisible by 3. Let the remainder of the number when divided by 9 be $r$, then, as we know, the sum of its digits also gives a remainder of $r$ when divided by 9, and thus the number formed by adding the digits also gives the same remainder when divided by 9 as $2 r$, and the sum of its digits also gives the same remainder. If we add the sum of the digits to the resulting number again, the number thus obtained gives the same remainder when divided by 9 as $4 r$. On the other hand, this number can be obtained from the sought number by swapping the digits, so it gives the same remainder when divided by 9 as $r$, i.e., $4 r$ gives a remainder of $r$ when divided by 9, so $3 r$ is divisible by 9, $r$ is divisible by 3, and thus the sought number is also.

On the other hand, we look for bounds on the digits of the number. Denoting the sought number by $10 x+y$, the number formed by the double addition, which can be written by swapping the digits, is $10 y+x$, and the increase is

$$
10 y+x-(10 x+y)=9(y-x)
$$

This is divisible by 9 on the one hand, and positive on the other, so $x<y$. This increase is the result of adding 4 digits to the sought number, among which there are at least two different ones ($x$ and $y$), so their sum can be at most $3 \cdot 9+8=35$, so

$$
9(y-x) \leq 35, \quad \text { so } \quad y-x \leq 3
$$

Thus, for the digits of the sought number,

$$
x<y \leq x+3
$$

Two-digit numbers that are divisible by 3 and have this property are the following:

$12,24,36,45,57,69,78$.

Testing these, it turns out that 12 and 69 are the solutions to the problem.