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## 'Negative Power' printed from http://nrich.maths.org/

We had a number of good answers, with commentary, on this problem including from Samantha (The Steele School), and Helen (Lady Margaret School).

Any calculation like this is the same as the bottom number to the power of all the other numbers multiplied together.

So all I needed to try out was :

$-4 \text{ to the power of }(-3 \times-2 \times-1)$ which is $-(4 \text{ to the power of } -6)$

$-3 \text{ to the power of } (-4 \times-2 \times-1)$ which is $-(3 \text{ to the power of } -8)$

$-2 \text{ to the power of } (-3 \times-4 \times-1)$ which is $-(2 \text{ to the power of } -12) $

$-1 \text{ to the power of } (-2 \times-3 \times-4)$ which is $-(1 \text{ to the power of } -24) $

That's each number as the bottom and the rest making a product to be the power or index number.

I noticed that having $4$ or $2$ at the bottom gives the same end result : $4$ to the power of $-6$ is the same as $2$ to the power of $-12$, which made me think a bit. I think it happens because $4$ is $2$ squared, and having $4$ at the bottom is like having only $2$ at the bottom and an extra factor of $2$ in the product that makes the power. But swapping the $2$ at the bottom and putting the
$4$ in the power stack also makes the product bigger by a factor of $2$.

So my four calculations, left as fractions, came out like this :

$$1/4096, 1/6561, 1/4096, 1$$

So $1$ is the largest result (putting $1$ at the bottom) and $1/6561$ is the smallest result (putting $3$ at the bottom).

Thank-you to everyone who sent in their results and ideas.