Dear Uncle Colin,

I noticed that $\frac{987654321}{123456789} \approx 8.0000000729$ in base 10. On further investigation, $\frac{321}{123}\approx 2$ in base 4, $\frac{654321}{123456}\approx 5$ in base seven and a similar pattern recurs in any base: $\frac{[b-1][b-2]\dots 54321}{12345\dots[b-2][b-1]}$ (where $[a][b]$ means the concatenation of digits $a$ and $b$) is very close to $b-2$. Any idea why?

Runs Are Tremendously Interesting, Oooya!

Hi, RATIO, and thanks for your message! The answer lies in the binomial expansion.

### In base 10

Let’s look at the base 10 version first, and then generalise it.

The nugget here is that $\frac{1}{81} = 0.\dot 01234567\dot 9$ - the nice pattern is slightly spoilt by a carry from the 10. This comes from $(1-x)^{-2} = 1 + 2x + 3x^2 + \dots$; evaluated at $x=\frac{1}{10}$, you get $\frac{100}{81} = 1.2345\dots$, and you can get to $\frac{1}{81}$ - or, more pertinently, something very close to 123456790 - from there. In fact, $\frac{10^{10}}{81} = 123,456,790.\dot 123456789\dot0$, which is only one-and-a-bit more than the denominator.

Let’s also think about (for reasons that will become clear) the sum of the numerator and denominator, which is 1,111,111,110. You might recognise an almost-pattern there, too; it looks very much like a ninth. In fact, it’s one and a smidge smaller than $\frac{10^{10}}{9}$.

Recap: 123,456,789 is about $\frac{x}{81}$ and 123,456,789 + 987,654,321 is about $\frac{x}{9}$, with $x =10^{10}$. That means 987,654,321 is roughly $\frac{8x}{81}$.

And voila: $\frac{987,654,321}{123,456,789} \approx \frac{8x/81}{x/81} = 8$.

## In base $b$

A similar pattern holds in general, only replacing 81 with $(b-1)^2$, 9 with $(b-1)$ and $10^{10}$ with $b^b$.

More concretely:

• $123\dots[b-2][b-1] \approx \frac{b^b}{(b-1)^2}$
• $[b-2][b-1]\dots321 + 123\dots[b-2][b-1] = 111\dots10 \approx \frac{b^b}{b-1}$
• So $[b-2][b-1]\dots321 \approx \frac{(b-2)b^b}{(b-1)^2}$
• And $\frac{[b-2][b-1]\dots321}{123\dots[b-2][b-1]} \approx b-2$.

Hope that helps!

- Uncle Colin