A 7-divisible number $K$ in binary form is as follows:

$$
K=10110101010101 x y z 110
$$

Determine the digits $x, y, z$.
Let's express $K$ in terms of its place values!

$$
\begin{gathered}
K=2^{19}+2^{17}+2^{16}+2^{14}+2^{12}+2^{10}+2^{8}+2^{6}+x \cdot 2^{5}+y \cdot 2^{4}+z \cdot 2^{3}+2^{2}+2^{1}= \\
=2 \cdot 8^{6}+6 \cdot 8^{5}+5 \cdot 8^{4}+2 \cdot 8^{3}+5 \cdot 8^{2}+8 \cdot(4 x+2 y+z)+6
\end{gathered}
$$

We will prove by complete induction that $7 \mid 8^{n}-1$. If $n=0$, then this is true. Suppose it is true for $n-1$; then

$$
8^{n}-1=8 \cdot 8^{n-1}-1=8 \cdot\left(8^{n-1}-1\right)+7
$$

By the induction hypothesis, $7 \mid 8^{n-1}-1$, hence $7 \mid 8 \cdot\left(8^{n-1}-1\right)+7$, so the statement is true for $n$.

Using our observation,

$$
\begin{gathered}
K=2\left(8^{6}-1\right)+6\left(8^{5}-1\right)+5\left(8^{4}-1\right)+2\left(8^{3}-1\right)+5\left(8^{2}-1\right)+ \\
+7(4 x+2 y+z)+3 \cdot 7+(4 x+2 y+z+5)
\end{gathered}
$$

is exactly divisible by 7 if

$$
4 x+2 y+z+5
$$

is divisible by 7. Since $5 \leq 4 x+2 y+z+5 \leq 12$, it must be that $4 x+2 y+z=2$. Since $x, y, z$ can only be 0 or 1, $x=0$. Furthermore, $z$ must be even, so $z$ is also 0, and finally $y=1$. Therefore, the problem has one solution: $x=0, y=1, z=0$.

Imre Ehreth (Bonyhád, Petőfi S. Gymnasium, 1st grade)

Note. Many submissions calculated as if the last digit (0) of $K$ in binary representation was on the binary place value. This is of course not the case, when writing whole numbers, the unit is the smallest place value.