1089 Trick

Theorem
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 $10^2a+10b+c$.

Its reverse is $10^2c+10b+a$

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 reverse, you get:
 * $9 \times 100 + 2 \times 9 \times 10 + 9 = 1089$

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.