1089 Trick

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

• 1986: David Wells: Curious and Interesting Numbers ... (previous) ... (next): $1089$
• 1995: Terry Wiley: Surreal School Stories: 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.

• 1997: David Wells: Curious and Interesting Numbers (2nd ed.) ... (previous) ... (next): $1089$
• 2002: David Acheson: 1089 and all that