1089 Trick

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Classic Puzzle

Take a three-digit number (one where the first and last digits are different by at least $2$).

Reverse it, and get the difference between that and the first number.

Reverse that, and add that to the difference you calculated just now.

You get $1089$.


Proof

Let the number you started with be expressed in decimal notation as $[abc]_{10}$.

Then it is $10^2 a + 10 b + c$.

Its reversal is:

$10^2 c + 10 b + a$

The difference between the two is $99a - 99c$ (or $99c - 99a$, it matters not).

This is a multiple of $99$.

The three-digit multiples of $99$ are:

$198$
$297$
$396$
$495$
$594$
$693$
$792$
$891$

By adding any one of these to its reversal, you get:

$9 \times 100 + 2 \times 9 \times 10 + 9 = 1089$

$\blacksquare$


Note: You need to make sure the difference between the first and last digits of the number you started with is at least $2$ so as to make sure that the first difference you calculate is definitely a $3$-digit number.


Sources