638*. Find all exact powers of the number 2 that occur among numbers of the form $6 n+8(n=0,1,2, \ldots)$.
$\triangle$ Let's write the equality

$$
2^{k}=6 n+8
$$

For which natural $k$ can the power $2^{k}$ be represented as $6 n+8?$

Calculations show that the values $k=3,5,7$ work, while the values $k=2,4,6,8$ do not. This suggests that, perhaps, all odd values of $k$, starting from 3, satisfy the condition of the problem, while all even values do not. Let's consider two cases.

1) Let $k$ be odd: $k=2 a+1(a \in N)$. Express $n$ in terms of $k$ from the equation:

$$
\begin{gathered}
2^{k}=6 n+8, \quad 6 n=2^{k}-8, \quad 3 n=2^{k-1}-4=\left(2^{k-1}-1\right)-3 . \\
n=\frac{1}{3}\left(2^{k-1}-1\right)-1=\frac{1}{3}\left(2^{2 a}-1\right)-1=\frac{1}{3}\left(4^{a}-1\right)-1 .
\end{gathered}
$$

Since the difference $4^{a}-1$ is divisible by 3 for any natural $a$ (see § 7, first divisibility), such a natural $n$ exists for any odd $k>1$. This means that all odd $k$, greater than 1, work.

2) Let $k$ be even: $k=2 a(a \in N)$. Then we get:

$$
2^{2 a}=6 n+8, \quad\left(2^{a}\right)^{2}=(6 n+6)+2, \quad\left(2^{a}\right)^{2}=3(2 n+2)+2
$$

But the square of a natural number, when divided by 3, cannot give a remainder of 2 (see the statement of problem 163 from § 5). The obtained contradiction shows that all even $k$ do not work.

Answer: powers of 2 with any odd natural exponent, greater than 1.