WTF!!! (a working mind reader)

That ain't no magic, it's an old trick. The result is always a number that, when the individual digits are added together, will result in a 9. If you check the grid on that web page, you'll notice that all such numbers have the same symbol:
 
I found it... :(

Too good to be true. The result is always a 9x multiplication result. And the second number to substract doesn't play any role.
 
Yeah I know. Wasn't very attentive on my first try.
There are loads of these tricks...

I call them mathematical passatempos...
 
Damn, I hit submit before I was done! Anyway, to continue, in the grid, the numbers 9, 18, 27, 36, 45, 54, 63, 72, 81 have the same symbol. No matter which two dogot number you pick and to the arithmetic on, you WILL end up with that same symol.
 
mubs said:
Damn, I hit submit before I was done!
Heheh, don't worry. From the time stamps of our posts I declare it a draw! :wink:
Do you know any other sites with stuff like this?
 
Naw. But if you go to StorageReview, in the Bar&Grille forum, Davin's thread ("impressionable you idiot" or something like that in the title) has some links that are fascinating--optical illusions. Check them out!
 
The more interesting question to me is: why does the arithmetic always give you a number which is a multiple of 9?

Joe
 
So the question is, why does the answer to the game always equal a multiple of the number nine.

If you write the game out in variable form we get:
Code:
(10^1A + 10^0B) - A - B = C

Which simplifies to:
Code:
10A + B - A - B = C

You can immediately tell that whatever you pick for the ones digit does not matter and has no bearing on the problem.

The previous answer then reduces to:
Code:
10A - A = C

9A = C

9 = C/A

And we see that no matter what number we pick for the tens position and no matter what number we get for the final result, their ratio will always equal nine. Or IOW, variable C will always be nine times the tens digit (variable A) in the number you picked.

Indeed since the symbol pattern is only predicated on the final result being a multiple of the number nine, the equation shows that the variable A can be any number. Unless the initial rules are changed, variable A may be, for example, a four digit number (thus resulting in an initial five digit number) and the equation must obey.
 
Thank you Clipton. Even though your proof is so neat and after the fact apparent, the result still seems a little magical.

Joe.
 
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