1. Determine all four-digit numbers $n=\overline{a b c d}$ such that inserting a digit 0 in any position results in a multiple of 7.
Solution. We start by observing that the number resulting from inserting a 0 at the end of $n$ is $10 n$, which, being a multiple of 7, forces $n$ to also be a multiple of 7. In fact, the following five numbers are multiples of 7:

$$
\begin{aligned}
n=\overline{a b c d} & =1000 a+100 b+10 c+d \\
x=\overline{a 0 b c d} & =10000 a+100 b+10 c+d \\
y=\overline{a b 0 c d} & =10000 a+1000 b+10 c+d \\
z=\overline{a b c 0 d} & =10000 a+1000 b+100 c+d \\
w=\overline{a b c d 0} & =10000 a+1000 b+100 c+10 d
\end{aligned}
$$

Since $n, x$ are multiples of 7, their difference $x-n=9000 a$ is also a multiple of 7, and since 9000 is not a multiple of 7, $a$ must be a multiple of 7. Since $n$ is a four-digit number, $a \neq 0$ and it must be $a=7$.

Similarly, $y-n=9000 a+900 b$ is a multiple of 7, and we know that $a$ is a multiple of 7, so $900 b$ is a multiple of 7. Since 900 is not a multiple of 7, $b$ must be a multiple of 7, and it follows that $b=0$ or $b=7$.

Analogously, reasoning with $z-n=9000 a+900 b+90 c$ we obtain that $c=0$ or $c=7$, and inserting the conditions that $a, b, c$ are multiples of 7 into the original number $n$ we deduce that $d$ is also 0 or 7.

Thus, the numbers sought are all those that start with 7 and the remaining digits are 0 or 7:

$7000,7007,7070,7077,7700,7707,7770,7777$.