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



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<a 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).


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)

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