1. A certain four-digit natural number is divisible by seven. If we write its digits in reverse order, we get a larger four-digit number, which is also divisible by seven. Moreover, when divided by 37, both mentioned four-digit numbers give the same remainder. Determine the original four-digit number.
1. Let the desired number be $n=\overline{a b c d}=1000 a+100 b+10 c+d$ and the number with the digits in reverse order be $k=\overline{d c b a}=1000 d+100 c+10 b+a$. Since both numbers $k, n$ give the same remainder when divided by 37, their difference is

$$
\begin{aligned}
k-n & =(1000 d+100 c+10 b+a)-(1000 a+100 b+10 c+d)= \\
& =999(d-a)+90(c-b)=37 \cdot 27(d-a)+90(c-b)
\end{aligned}
$$

divisible by the number 37, from which $37 \mid 90(c-b)$. Since 37 is a prime number and the number 90 is not its multiple, it must be that $37 \mid c-b$. This is possible for digits $b, c$ only in the case when $b=c$. Conversely, if $b=c$, it follows from the expression (1) that regardless of the values of the digits $a, d$, the difference $k-n$ is divisible by the number 37, so the numbers $n$ and $k$ indeed give the same remainder when divided by 37. We can therefore assume that $n=\overline{a b b d}$ and $k=\overline{d b b a}$, and further consider only the conditions of the problem about divisibility by seven.

From the conditions $7|n, 7| k$ it follows that

$$
7 \mid k-n=37 \cdot 27(d-a)
$$

(by substituting $b=c$ into expression (1)), from which, given the coprimality of the number 7 with the product $37 \cdot 27$, we get that $7 \mid d-a$. Since according to the problem statement $k>n$, it also holds that $d>a$; such digits $d$, $a$ satisfy the condition $7 \mid d-a$ only in the case when $d-a=7$. Specifically, then either $a=1$ and $d=8$, or $a=2$ and $d=9$. (The possibility $a=0, d=7$ is excluded because $a$ is the first digit of the four-digit number $n$.)

In the case $a=1, d=8$, the numbers

$$
\begin{aligned}
n & =\overline{1 b b 8}=1008+110 b=7 \cdot(144+15 b)+5 b, \\
k & =\overline{8 b b 1}=8001+110 b=7 \cdot(1143+15 b)+5 b
\end{aligned}
$$

are divisible by seven if and only if $7 \mid 5 b$ or $b \in\{0,7\}$. This gives us the first two solutions $n=1008$ and $n=1778$.

Similarly, in the case $a=2$ and $d=9$, when

$$
\begin{aligned}
& n=\overline{2 b b 9}=2009+110 b=7 \cdot(287+15 b)+5 b, \\
& k=\overline{9 b b 2}=9002+110 b=7 \cdot(1286+15 b)+5 b,
\end{aligned}
$$

the same condition $7 \mid 5 b$ leads to the remaining two solutions $n=2009$ and $n=2779$.

The desired four-digit number is any of the numbers 1008, 1778, 2009, 2779 (and no others).

Alternative solution. Retain the notation of the numbers $n, k$ and their digits from the original solution. If we start the solution by analyzing the conditions $7 \mid n$ and $7 \mid k$, considering the remainders of the powers $10^{3}, 10^{2}$, $10^{1}, 10^{0}$ when divided by seven (which are 6, 2, 3, 1, respectively), we find that these conditions can be simplified to the form

$$
7 \mid 6 a+2 b+3 c+d, \quad \text { resp. } \quad 7 \mid 6 d+2 c+3 b+a \text {. }
$$

By adding both expressions, we get $7 \mid 5(b+c)$ or $7 \mid b+c$. If we find, as in the original solution, that $37 \mid k-n \Leftrightarrow b=c$, from the condition $7 \mid b+c$ it follows that $b=c \in\{0,7\}$; from the remaining two conditions $7 \mid 6 a+d$ and $7 \mid 6 d+a$, which are both equivalent to one relation $7 \mid d-a$, and considering $d>a>0$, we find both suitable pairs of digits $(a, d)=(1,8)$ and $(a, d)=(2,9)$.

For a complete solution, award 6 points, of which 2 points for establishing the condition $b=c$, 2 points for deriving the relation $d-a=7$, 1 point for justifying $7 \mid b(=c)$ or for finding $7 \mid b+c$, 1 point for correctly listing all four solutions.