SOLUTION. Let's denote the digits of Dana's number as $a, b, c$. The information about the remainders when divided by seven from the problem can be rewritten into equations

$$
\begin{aligned}
& 100 a+10 b+c=7 x+2, \\
& 100 b+10 a+c=7 y+3, \\
& 100 a+10 c+b=7 z+5 .
\end{aligned}
$$

The digits $a$ and $b$ cannot be zero, because both the first and the second number in the problem are three-digit numbers; thus, $a, b \in\{1,2, \ldots, 9\}, c \in\{0,1,2, \ldots, 9\}$ and $x, y$ and $z$ are integers.

Now we will try to find the remainders of the digits $a, b, c$ when divided by seven. This will give us at most two possibilities for each digit. Looking at the coefficients in equations (1)-(3), we see that by a suitable subtraction, we can eliminate two digits $a$ and $c$ at once if we subtract equation (3) from ten times equation (2). We will then simplify the result step by step to find the remainder of the digit $b$ when divided by seven

$$
\begin{aligned}
10(100 b+10 a+c)-(100 a+10 c+b) & =10(7 y+3)-(7 z+5), \\
999 b & =70 y-7 z+25, \\
5 b & =70 y-7 z-7 \cdot 142 b+7 \cdot 3+4, \\
15 b & =3(70 y-7 z-7 \cdot 142 b+7 \cdot 3)+3 \cdot 4, \\
b & =3(70 y-7 z-7 \cdot 142 b+7 \cdot 3)-7 \cdot 2 b+12 .
\end{aligned}
$$

Since all terms on the right side of the last equation are divisible by seven except for the number 12, $b$ gives the same remainder when divided by seven as the number 12, and the only suitable digit $b$ is $b=5$.

By subtracting equation (3) from equation (1), we get the equation $9 c-9 b=7(z-x)+3$, and after substituting $b=5$ we get

$$
\begin{aligned}
9 c-9 \cdot 5 & =7(z-x)+3, \\
2 c & =7(z-x-c)+48, \\
8 c & =4 \cdot 7(z-x-c+6)+4 \cdot 6, \\
c & =4 \cdot 7(z-x-c+6)-7 c+7 \cdot 3+3 .
\end{aligned}
$$

From the divisibility of individual terms by seven, we get $c=3$.

Finally, by substituting $b=5$ and $c=3$ into the first equation, we can easily calculate the value of $a$:

$$
\begin{aligned}
100 a+10 b+c & =7 x+2, \\
100 a+53 & =7 x+2 \quad(53=7 \cdot 8-3,98=7 \cdot 14), \\
2 a & =7 x+2-7 \cdot 8+3-7 \cdot 14 a, \\
a & =4(7 x-7 \cdot 8-7 \cdot 14 a)+4 \cdot 5-7 a,
\end{aligned}
$$

from which it follows that the digit $a$ gives the same remainder when divided by seven as the number $20=7 \cdot 2+6$, and thus $a=6$.

Dana therefore wrote the number 653 on the paper, so the number formed by swapping the first and last digit is $356=7 \cdot 50+6$ and gives a remainder of 6 when divided by seven.
4. Dana wrote a three-digit number on paper, which when divided by seven gives a remainder of 2. By swapping the first two digits, a three-digit number was formed that when divided by seven gives a remainder of 3. The number formed by swapping the last two digits of the original number gives a remainder of 5 when divided by seven. What remainder will the number, which is formed by swapping the first and last digits of Dana's number, have when divided by seven?