Numerical Reverser

How am I psychic? I'm not. I'm a mathematician! This is a rather simple trick and the math behind it isn't particularly complex.

You are asked to pick a two digit number. Any two-digit number can be expressed as:

10x + y

Where x and y are positive integers between 0 and 9. For example, the number 65 is 60 + 5 and the number 38 is 30 + 8.

The next step in the trick is to reverse the digits, which yields

10y + x

And then to add the original number:

10y + x + 10x + y
=11x + 11y

The trick to all of this is that the above expression is cleanly divisible by eleven. To finish the trick, you must know a property of multiples of eleven: the alternating sum of the digits is zero. That means that the first digit, minus the second digit, plus the third digit, minus the fourth digit (and so forth) will yield 0 if the number is divisible by eleven.

So how does that help us? Well, the first thing to note is that 11x + 11y (the number plus its reverse) is never going to be greater than 198. So if the total is three digits long, the first digit must be a one. Secondly, the sum is divisible by eleven. Given that, you know enough to work backwards from the information to find the other numbers.

Here's an example. Suppose the number is three digits long and ends in three. We know that the first digit is one (as I just said above) and the last digit is three. Furthermore, the number as a whole has an alternating sum of zero. So, if our number looks like 1?3, we know that 1 - ? + 3 = 0, or 4 - ? = 0, or ? = 4. So the number is 143.

With a bit of practice, this trick becomes extremely simple.