1089 Magic Trick

[grsbanks]

For a quick mathematical explanation of this... A 3-digit number "abc" is basically:

\(100a+10b+c\) where each of the terms a, b and c is a natural number from 0 to 9 inclusive.

Reversing the digits gets you a new number, which is:

\(100c+10b+a\)

If you want the difference between the two, there are two cases. Either \(a>c\) or \(a<c\). As has already been pointed out, this trick doesn't work if \(a=c\) because the original number and its reverse are the same, thus cancelling out the whole thing.

First scenario: \(a>c\)

\(100a+10b+c-(100c+10b+a)=99a-99c=99(a-c)=(100-1)(a-c)=100(a-c)+(c-a)\)
\(100(a-c)+(c-a)=100(a-c-1)+100+(c-a)=100(a-c-1)+90+(10+c-a)\)

We know that \(a>c\), so the expression \((10+c-a)\) is necessarily going to be less than 10 and greater than or equal to 1, i.e. it's going to be a single digit. Similarly, \(0 \leqslant (a-c-1) \leqslant 8\).

We therefore have the 3 digits of the difference between the original number and its reverse. Let's get the reverse of this difference by reversing the units and hundreds digits:

\(100(10+c-a)+90+(a-c-1)\)

If we now add the difference to its reverse, we get:

\(100(a-c-1+10+c-a)+180+(10+c-a+a-c-1)=100\times9+180+9=1089\)

Second scenario: \(a<c\)

Just start with the reverse of the original number and you'll find yourself in the first scenario again.