Example 10. (IMO6-1(1)) Find all positive integers $n$ such that $2^{n}-1$ is divisible by 7.
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When $n=3 k$, $2^{n}-1=2^{3 k}-1=8^{k}-1$ \\
$=(7+1)^{k}-1 \equiv 0(\bmod 7)$; \\
When $n=3 k+1$, $2^{n}-1=2^{3 k+1}-1$ \\
$=2(7+1)^{b}-1 \equiv 1(\bmod 7)$; \\
When $n=3 k+2$, $2^{n}-1=2^{3 k+2}-1$ \\
$=4(7+1)^{k}-1 \equiv 3(\bmod 7)$. \\
Thus, only when $n$ is a multiple of 3, $2^{n}-1$ is divisible by 7.
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