I’ve got a new favourite party trick. It goes like this:

“Pick a number ((If you’re talking to a mathematician, you may need to translate this as ‘Let $x$ be an integer such that $1 < x < 10.$’)) between 1 and 10 - don’t tell me what it is. Pick another number between 1 and 100 - you can tell me that one. Now work out the first number to the power of the second for me on this handy calculator, and I’ll tell you the first number.”

For example, you might pick your first number, tell me your second number is 63 and that the calculator’s given you something weird as an answer, something like $1.084 \times 10^{44}$.

I reassure you, that’s totally ok, and that your original number was 5.


How on earth did I know that? Well, clearly, I’ve memorised the 63rd powers of all of the possible numbers… it’s the only possible way.

Unless I did something sneaky with logs, of course. I may have alluded to it in this apology to base-10 logarithms. But here’s what I did:

I took logs base 10, very roughly, of the big number you gave me. $\log_{10}(1.084 \times 10^{44})$ is something in the neighbourhood of 44 ((Precisely, it’s 44.035.)). Certainly not far off. Standard form makes that sort of thing easy.

Then I divided it by the smaller number you gave me. $\frac{44}{63}$ isn’t the easiest thing to do in your head, but it’s only a tiny bit short of $\frac{45}{63}$ which cancels to $\frac{5}{7}$ – which we all know is 0.714285 (recurring).

Being a mathematical ninja, I know my logs base 10 – and in particular, $\log_{10}(5)$ is about 0.7, while $\log_{10}(6)$ is about 0.78. The 0.714 was an overestimate, so I go for the lower (and closer) of the two – and say 5 as quickly as I can.

To do this trick yourself, you’ll need to know the log tables from the other post – roughly is good enough – and be able to approximate fractions in your head, which is a trickier skill… but one you can work on, right? After a while, you’ll start to spot patterns that make the numbers easier.

* Edited 21/7/2014 to fix LaTeX and footnotes.