elementary number theory - If $n$ is an odd integer prove that $n - 2^k$ is divisible by $3$

So let $n$ be a odd integer. Show that $n - 2^k$ is divisible by $3$ if $k$ is SOME SPECIFIC positive integer. $k \ge 0$. So there only has to exist one. For example:

$$7 - 2^2 = 3$$ is divisible by $3$

The approach is modular arithemetic, but it is hard since,

$$2 \equiv 2 \pmod{3}$$

$$n \equiv p \pmod{3}$$

It is hard to combine these? What should I do?

Answer

Hint $\ {\rm mod}\ 3\!:\ 2^2\equiv 1\,$ so $\,2^k\equiv 2^0\equiv 1$ or $\,2^k\equiv 2^1 \equiv 2$. Thus $\,n\equiv 2^k\iff 3\nmid n$