A nice arithmetical trick to show to children is the 1089 trick. It’s not really a trick, just a surprising and cool arithmetical result that appears to be inexplicable at first sight. And a bit spooky. What is it about 1089 that is so universally significant!? Here’s the trick…
Take any 3 digit number for which the first and last digits differ by 2 or more, so for example, take 247. Then reverse the number and subtract so :
742 – 247 = 495
Reverse this result and add, so:
495 + 594 = 1089
This answer, 1089, will result from any 3 digit number where the first and last digits differ by at least 2.
The explanation becomes clear when you look at the algebra behind the sums.
Take any 3 digit number, abc. Writing the subtraction (this time taking the reversed number from the original – it doesn’t matter which way round it goes) gives:
abc – cba = 100a -100c – a – c (the b’s in the tens column cancel out)
= 99a – 99c = 99(a-c)
Since a and c differ by at least 2, the possible values of a-c are: 2,3,4,5,6,7,8
So, 99(a-c) must equal one of the following: 198, 297, 396, 495, 594, 693, 792
Now say that the result of abc – cba is xyz
Then the end result, e, of adding xyz to zyx can be written as 100x + 10y + z + 100z + 10y + x
e = 100(x+z) + 20y +x + z
But from our list of possible values of 99(a-c), we can can see that the middle digit, y, is always 9. Also, the addition of the two end digits is always 9. So x+z = 9.
So our final sum e = 100(9) + 20(9) + 9
e= 900 + 180 + 9
e=1089
Always.
QED!
Phew! And that’s a simple proof by any mathematical standard.
