Puzzle - 1089 Trick
- Think of a three digit number \(abc\) where \(a - c\geq 2\)
- Subtract the number with the digits reversed, i.e. \(cba\)
- Add on the reverse of the result
- The answer is \(1089\)
Proof
Result of steps 1-2 is \([a-c-1][9][10+c-a]\) where the digits are in square brackets.
Moving onto step 3, we have:
\[\begin{align*} &[a-c-1][9][10+c-a] + [10+c-a][9][a-c-1] \\ &= [a-c-1+10+c-a][9+9][10+c-a+a-c-1]\\ &= [a-c-1+10+c-a+1][8][9]\\ &= 1089\\ \end{align*}\]