Prove that $m$ is an integer Suppose $n$ is a odd integer. It satisfies:
$$3^{s} < n < 3^{s+1}$$ For some integer $s \ge 0.$
Show that:
$$m = \frac{n - 3^{s}}{2}$$ Is an integer.
So,
$$2m = n - 3^{s}$$
But that wont help? How to approach this?
 A: As $n$ is odd, and $3^s$ is a power of $3$ (and hence odd), then their difference is even.
A: The problem statement contains many red herrings. Note that
$$
\frac{n-3^s}{2}
$$
is an integer if 
$$
n-3^s
$$
is even. Can you prove that $n-3^s$ is even?
A: If you're given that $n$ is odd, then the original inequality implies that $n-3^s$ must be a positive even integer since all powers of 3 are odd. We can therefore divide by 2 and still have an integer.
A: The difference of two odd integers is even: $(2j + 1) - (2k + 1) = 2(j - k)$.
We're told $n$ is odd. We're also told $s$ is an integer, which means that $3^s$ is an odd integer. Therefore $n - 3^s = 2m$ with $m$ also being an integer. Then $\frac{2m}{2} = m$, which, as we just said, is an integer.
All the other details are just a way to dress up a very simple problem to look more complicated than it actually is.
A: I really don't like $s$ as an exponent, when it gets to a certain size, it looks more like an asterisk than a letter. I'll use $k$ instead.
$n$ is odd and between two powers of $3$. This means that $n - 3^k$ is even and positive, and therefore we can divide it by $2$ to obtain a positive integer (which may be odd or even, I don't know, couldn't say).
There, you're done.
But humor me, let's work out an example. $n = 47$, so $k = 3$. Then $$m = \frac{47 - 27}{2} = 10.$$
