# Impossible Sums - Proof of Converse

##### Age 14 to 18Challenge Level

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).