239*. Find all natural $n$ for which the difference $2^{n}-1$ is divisible by 7.
$\triangle$ Let's perform an enumeration, assigning $n$ the values $1,2,3,4,5,6$ and 7. It turns out that the difference $2^{n}-1$ is divisible by 7 only when $n=3$ and $n=6$. It seems that this difference is divisible by 7 only when $n$ is divisible by 3. Let's test our hypothesis. For this, we will consider three cases.

1) Suppose $n$ is divisible by 3: $n=3k (n \in N)$. Then

$$
2^{n}-1=2^{3k}-1=8^{k}-1
$$

and $8^{k}-1$ is divisible by 7 for any natural $k$.

2) Suppose $n$ gives a remainder of 1 when divided by 3: $n=3k+1(k=0,1,2, \ldots)$. We get:

$$
\begin{aligned}
& 2^{n}-1=2^{3k+1}-1=2 \cdot 2^{3k}-1=2 \cdot 8^{k}-1= \\
& \quad=2\left(\left(8^{k}-1\right)+1\right)-1=2\left(8^{k}-1\right)+1
\end{aligned}
$$

The first term of the obtained sum is divisible by 7 for any non-negative integer $k$, while the second term is not, so the entire sum is not divisible by 7.

3) Suppose $n$ gives a remainder of 2 when divided by 3: $n=3k+2(k=0,1,2, \ldots)$. Similarly to the previous case, it can be shown that in this case the difference $2^{n}-1$ is also not divisible by 7.

Answer: all $n$ that are divisible by 3.