The problem Impossible Sums asks you to think about which numbers can be written as a sum of consecutive numbers, and which numbers cannot.
Below is a proof that any number which is not a power of $2$ can be written as a sum of consecutive positive numbers.
The proof has been scrambled up. Can you rearrange it back into its original order?
You might find it helpful to take a look at the example below before rearranging the proof. You might also want to refer to it whilst you are rearranging the statements.
For example, consider the number $22$. This can be written as: $$22 = 11 \times 2=2+2+2+2+2+2+2+2+2+2+2$$
This can be replaced with a set of 11 consecutive numbers centered around 2:
$$-3,-2,-1,\phantom{-}0,\phantom{-}1,\phantom{-}2,\phantom{-}3,\phantom{-}4,\phantom{-}5,\phantom{-}6,\phantom{-}7$$
and a quick check shows that these add to give $22$.
Cancelling the negative numbers with their positive equivalents gives:
$$22= \color{red}{\bf (-3)}+\color{red}{\bf (-2)}+\color{red}{\bf (-1)}+0+\color{red}{\bf 1}+\color{red}{\bf 2}+\color{red}{\bf 3}+4+5+6+7$$
$$\implies 22 = 4+5+6+7$$
In this case we have $22=11 \times 2 = (2\times 5 + 1 ) \times 2$.
This is of the form $(2k+1)m$ as appears in the proof below with $k=5$ and $m=2$. To find the $11$ consecutive numbers we considered the $5$ numbers before $2$ and the $5$ numbers after $2$ (as well as $2$ itself).