6. (IMO-5 Problem) (1) Find all positive integers $n$, such that $2^{n}-1$ is divisible by 7; (2) Prove that for any positive integer $n, 2^{n}+1$ cannot be divisible by 7.
6. (1) Notice that $7=(111)_{2}, 2^{n}-1=(\underbrace{11 \cdots 11}_{m(1)})_{2}$. Clearly, the required conclusion holds if and only if the bit length of the latter is an integer multiple of the bit length of the former or the dividend is zero, i.e., there should be $n=3 k$ ($k$ is a non-negative integer) as the solution.
(2) Notice that $2^{n}+1=(1 \underbrace{00 \cdots 01}_{n \text { bits }})_{2}$. Observing the long division (omitted), the remainder obtained at each step of the division is always 1. Also, note the composition of the last digit. Therefore, when performing the final step of the division, the dividend unit can only be one of three cases: $(1001)_{2}, (101)_{2}, (11)_{2}$, none of which can be divided by $(111)_{2}$, hence the proof.