<s>[INST] 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? 
 [/INST] So if $n$ is an odd integer, you have to prove that there exists some integer exponent $k$ such that $3|(n - 2^k)$? Trouble is, it depends on what $n$ is. There are either infinitely many solutions or none at all.
Let's look at this modulo 6: there are three possibilities: $n \equiv 1, 3 \textrm{ or } 5 \pmod 6$. For the powers of 2, we have $2^k \equiv 2 \textrm{ or } 4 \pmod 6$ for $k > 0$. We want $n - 2^k \equiv 0 \textrm{ or } 3 \pmod 6$.
But how to combine all these? It's actually quite easy: we just subtract! Just be sure to mind the wrap-around.
For example, if $n \equiv 1 \pmod 6$ and $2^k \equiv 4 \pmod 6$, then $n - 2^k \equiv 1 - 4 \equiv 3 \pmod 6$.
But what if $n \equiv 3 \pmod 6$? No $k$ can work: if $2^k \equiv 2 \pmod 6$, then $n - 2^k \equiv 3 - 2 \equiv 1 \pmod 6$, and if $2^k \equiv 4 \pmod 6$, then $n - 2^k \equiv 3 - 4 \equiv 5 \pmod 6$. Don't forget $k = 0$, but that's of no help here: $n - 1 \equiv 3 - 1 \equiv 2 \pmod 6$.
The solution for $n \equiv 5 \pmod 6$ should be easy enough for you to figure out now.
One of the other answerers mentioned a requirement for $n - 2^k$ to be positive, but you didn't. I don't know the rules of the contest, so I assume negative values are perfectly acceptable. For example, $-9 \equiv 3 \pmod 6$ since $(-2) \times 6 + 3 = -12 + 3 = -9$.
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