1089
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:
- $9 \times 100 + 2 \times 9 \times 10 + 9 = 1089$
$\blacksquare$
Note: You need to make sure the difference between the first and last digits of the number you started with is at least two so as to make sure that the first difference you calculate is definitely a 3-digit number.
Sources
- Terry Wiley: Surreal School Stories (1995): No. 2 Vol. 1: The Late Seventies Masonettes: The Second Chapter: 'My Brain Hurts'. Jocasta Dribble has been passed the list of number codes for the secret society she has just joined. The code for "weird" is $1089$: "a funny number". Examples of its use ($234$, $742$ and $876$) are illustrated on the piece of paper.
