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Author Topic: neat number trick..  (Read 547 times)
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Dr.Jeckyl
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« on: December 11, 2007, 04:55:28 AM »

I can't seem to figure this one out.

http://www.quizyourprofile.com/guessyournumber.swf

If you figure it out let us know. Mike... Wink
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« Reply #1 on: December 11, 2007, 04:56:49 AM »

After 2 more tries I think I've figured it out. I'll post it tomorrow.
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farmboy
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« Reply #2 on: December 11, 2007, 05:14:17 AM »

yeah, i figured it out.  took me whipping out my TI to see if it could trick random numbers first, though.
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MaddGIJoe
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« Reply #3 on: December 11, 2007, 05:31:12 AM »

My guess at it
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Zazoo
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« Reply #4 on: December 11, 2007, 12:13:37 PM »

Yeah, there isn't any math to this one.
The one you posted a year or two ago was better Dr. J. Still have it?

~Mike
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« Reply #5 on: December 11, 2007, 08:47:52 PM »

I'll see if I can find it.
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« Reply #6 on: December 11, 2007, 08:51:52 PM »

http://www.milaadesign.com/wizardy.html
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Mutt_Dog
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« Reply #7 on: December 11, 2007, 11:39:42 PM »

Ok I'm all freaked out!
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Dr.Jeckyl
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« Reply #8 on: December 11, 2007, 11:59:39 PM »

I'll quote Mike since the thread isn't visible to everyone.

Quote from: Mike/Zazoo
Nah, you're not nuts. It's just a mathematical trick.

The trick can be explained algebraically if you know that you can expand any decimal number using powers of ten, like this:

Say you have the number 7,463.

You can write it in expanded powers of ten form this way:
7(10^3) + 4(10^2) + 6(10^1) + 3(10^0)

Since:
10^3 = 1000
10^2 = 100
10^1 = 10
and,
10^0 = 1

When you multiply it out you get:
7000 + 400 + 60 + 3
Which, when added, is 7463.

So…
The site asks you to pick any two digit number.
If you use a as the first digit (tens place), and b as the second digit (one’s place), with a and b each being a number between 1 and 9, you can write your two digit number in generic, expanded form:
a(10^1) + b(10^0)

(So if your number was 34, then a=3 and b=4, so 3(10^1) + 4(10^0) = 3(10) + 4(1) = 34)

Now the site asks you to add the digits of the number together, and then subtract this total from the original number. Mathematically you can represent this like:
a(10^1) + b(10^0) – (a + b) =

Which means:                     
(Original number): a(10^1) + b(10^0)
(minus): –                           
(the two digits of the number added together): (a + b)

Now work it algebraically…

Clear the second set of parentheses and you get:
a(10^1)+b(10^0) – a – b =

Simplify the powers of ten parts:
a(10) + b(1) – a – b =
10a + b – a – b

Combine like terms and the b’s cancel out leaving just:
9a

Since the first digit of your original number was a, and the original number had to be a two digit number (any number from 10 to 99), a has to be between 1 and 9. So no matter what number you pick in the beginning, the number that you get in the end will always be a multiple of 9 (and what multiple of 9 depends only on the value of the tens place digit (the a) of your original two digit number):
(9*1) = 9
(9*2) = 18
(9*3) = 27
(9*4) = 36
(9*5) = 45
(9*6) = 54
(9*7) = 63
(9*Cool = 72
(9*9) = 81

Look at the symbol for each multiple of nine on the last page:
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« Reply #9 on: December 14, 2007, 11:08:18 AM »

Being an accounting major we run into this a lot.

Any error made by transposition (switching numbers eg: 91 becomes 19) will be a factor of 9.
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