Expression of an Integer as a Power of 2 and an Odd Number (Chartrand Ex 5.4.2[a]) 
Let $n$ be a positive integer. Show that every integer $m$ with $ 1 \leq m \leq 2n $ can be expressed as $2^pk$, where $p$ is a nonnegative integer and $k$ is an odd integer with $1 \leq k < 2n$.

I wrote out some $m$ to try to conceive the proof. I observed: 
$\bbox[5px,border:1px solid grey]{\text{Case 1: $m$ odd}}$ Odd numbers $\neq 2p$, thus the only choice is to put $p = 0$ and $k = m$.
$\bbox[5px,border:1px solid grey]{\text{Case 2: $m$ even and power of 2}}$ Then $p$ can be determined by division or inspection to "square with" the power of $2$. This requires $k = 1$. Is an explicit formula for $p$ necessary?
$\bbox[5px,border:1px solid grey]{\text{Case 3: $m$ even and NOT A power of 2}}$ 
$\begin{array}{cc}
\\
\boxed{m = 6}: 6 = 2^1 \cdot 3 \qquad \qquad & \boxed{m = 10}: 10 = 2^1 \cdot 5 \qquad  \qquad & \boxed{m = 12}: 12 = 2^2 \cdot 3  \\
\boxed{m = 14}: 14 = 2 \cdot 7 \qquad \qquad & \boxed{m = 18}: 18 = 2^1 \cdot 9 \qquad  \qquad & \boxed{m = 20}: 20 = 2^2 \cdot 5\\
\boxed{m = 22}: 22 = 2 \cdot 11 \qquad \qquad & \boxed{m = 24}: 24 = 2^3 \cdot 3 \qquad  \qquad & \boxed{m = 26}: 26 = 2^1 \cdot 13
\end{array}$

Solution's Proof by Contradiction: $\color{#0073CF}{\Large{\mathbb{[}}}$Let $p$ be the greatest nonnegative integer
  such that $2^p | m. \color{#0073CF}{\Large{\text{]}}}$ $\color{red}{\Large{\text{[}}}$ Then $m= 2^pk$ for some positive integer $k$. Necessarily $k$ is odd,
  for otherwise this would contradicts the definition of $p$. $\color{red}{\Large{\text{]}}}$

$\Large{\text{1.}}$ Could someone please expound on the answer? It differs from my work above?
$\Large{\text{2.}}$ Is there a general formula/pattern for Case $3$? 
I referenced 1. Source: Exercise 5.42(a), P125 of Mathematical Proofs, 2nd ed. by Chartrand et al

I personally found this problem easier to do if you consolidate case 2 and 3 into one case and then prove by contradiction:
Assume m is even and that m  cannot be written in the form $2^p \bullet k$ where $p$ is a nonnegative integer and k is an odd integer with $1 \le k \le 2n$. Thus, $m \neq 2^p \bullet k$. We can rewrite $2^p \bullet k = 2(2^{p-1} \bullet k)$. (Note that because m is even $p \ge 1$ and so $p-1 \ge 0$). This implies that  m is not even, which directly contradicts our assumption that m is even. Therefore, we must conclude that m can be written in the form $2^p \bullet k$ where $p$ is a nonnegative integer and k is an odd integer with $1 \le k \le 2n$. 
