HAY GUYS

DeLorean · Local time: Jun 2, 2007, 11:12 PM

I know it's something really stupid and simple, but I can't seem to put my finger on how this works.

http://www.learnenglish.org.uk/games...er-central.swf

Jam it back in, in the dark.

Dullenplain · Local time: Jun 3, 2007, 12:58 AM

The trick goes like this.

1. Take any two-digit number: any will do (Ex: 11, 35, 67, 84)

2. Add the two digits together: (Ex: 2, 8, 13, 12)

3. Subtract the first number with the second: (Ex: 9, 27, 54, 72)

Now, note that no matter what two-digit number you chose, you will always end up with a number divisible by 9.

This is how the trick works. The chart the "gopher magician" puts up has symbols correlating with numbers, but note that all numbers that are 9 and its multiples (18, 27, 36, 45, 54, 63, 72, 81, 90, 99) have the exact same symbol.

Thus it is able to "predict" what symbol it is correctly every time.

There's nowhere I can't reach.

einherjar · 07:17 AM

If you need algebraic proof to convince you, here it is.

Let A and B be any two integers from 0-9, then 10A + B will be your chosen number where B is your ones digit and A is your tens digit.

Notice that subtracting the sum of the digits from the number will always give a multiple of 9, i.e.

10A + B - (A+B) = 9A where A is an integer from 0-9

That's how it's mathematically sound.

If you want to think about it intuitively, you can think of any two-digit number as being made up of 10 times the tens digit added to the ones digit. For example, 67 = 10 times of 6 + 7.

Since you subtract the sum of the ones digit and tens digit, it will not matter what your last digit is; you will end up subtracting it away anyway. Focus on your tens digit:

You realize that you have to minus it off from 10 times of it. That is, in the case of 67, 10 times of 6 - 6.

You will then be left with 9 times of the tens digit, which, in this example, is 9x6 = 54.

That's why you always get a multiple of 9. Note that this works for any number that has 2 digits or more too, because you can extend this reasoning to encompass further digits.

Hope that helped.

This thing is sticky, and I don't like it. I don't appreciate it.