## 1089 and all that

Here is the solution to Adam McBride’s problem.

What one needs to explain here is why the result is independent of the 3 digit number you start with.

A general 3 digit whole positive number has the form $\alpha \beta \gamma$, where $\alpha,\beta, \gamma$ are integers between 0 and 9. So let us start with some such number $abc$ and without loss of generality we can assume that $a>c$ (because $a=c$ is not allowed and if $c>a$, we anyway are going to turn the number around and subtract the smaller from the bigger one).

Now, really

$abc = 100a+10b+c.$

Turning this around, we have

$cba=100c+10b+a.$

Hence

$x= abc-cba = 100(a-c)+(c-a).$

It is not yet clear what the $\alpha \beta \gamma$ form of $x$ is, because $(c-a)$ is negative as $c by assumption. So let us do the oldest trick in the Book and add and subtract 1:

$x = 100(a-c-1+1)+(c-a)=100(a-c-1)+90+(10+c-a).$

So $x=\alpha \beta \gamma$, with $\alpha=a-c-1\geq 0$, $\beta=9$, and $\gamma=10+c-a \leq 9$. Now we can turn $x$ around. This gives

$y = 100(10+c-a)+90+ (a-c-1).$

Finally, all we have to do is add $x$ and $y$:

$x+y = 100(a-c-1+10+c-a)+90+90+(10+c-a+a-c-1) = 900+180+9 = 1089,$

and all the $a$ and $c$ have cancelled ($b$ got rid of long ago, in the computation of $x$).

By the way note that $x$ is necessarily divisible by 9 and 11 because $x=99(a-c)$.

(MG)

Postscript. Some of you may have recognised the title to this post as a reference to Sellar & Yeatman’s 1066 and All That, the classic account of school history as remembered years later. It’s also the title of a book by the remarkably versatile applied mathematician David Acheson — see this article on plus.maths.org to find out more. And speaking of Books… (DP)