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Bob_VT
12-19-2008, 09:03 AM
WHO IS YOUR ROLE MODEL???




Try it without looking at the answers......



please don't look down until you do it, you'll love it I promise



GET A CALCULATOR (YOUR COMPUTER HAS ONE ON IT)



1) Pick your favorite number between 1-9



2) Multiply by 3 then



3) Add 3, then again Multiply by 3 (I'll wait while you get the calculator....)



4) You'll get a 2 or 3 digit number….



5) Add the digits together



Now Scroll down ..............







































































Now with that number see who your ROLE MODEL is from the list below:



1. Einstein




2. Nelson Mandela




3. Abraham Lincoln




4. Helen Keller




5. Bill Gates




6. Gandhi




7. George Clooney




8. Thomas Edison




9. Bob_VT



10. Abraham Lincoln


Happy Holidays!!!:biggrin:

jclo3313
12-19-2008, 09:14 AM
Bob your my hero.:biggrin:

ChinoCharles
12-19-2008, 09:14 AM
LOL! I like it!

bobselectric
12-19-2008, 09:20 AM
Love it Bob!

Rain
12-19-2008, 09:22 AM
oh.. haha.
niceee!

ChinoCharles
12-19-2008, 09:32 AM
I'm waiting for someone to come on here and tell us they got Hellen Keller.

KCALB SIRAY
12-19-2008, 01:39 PM
sombitch, Bob! lol

SuperFunBall
12-19-2008, 02:20 PM
Hey Abe made the top ten twice

YAR1S
12-19-2008, 02:51 PM
hahaha nice. I've seen this before... I dodnt even have to go through it... I just know what happens, haha.

GeneW
12-20-2008, 04:09 AM
Let K be 1 to 9

1. 3K times three

2. 3k+3 add to that three

3. 3(3K+3) = 9(K+1) multiply the whole thing by three again and then group out the nine.

Let K+1 = J, where J is from 2 to 10

9J = 10a+b (in other words a representation of a decimal number in terms of powers of ten....

Turns out that all multiples of nine between two and ten have digits that "sum" to nine.

Tabular Form...

18 then a=1 and b=8
27 then a=2 and b=7
36 then a=3 and b=6
45 and so on....
54
63
72
81
90

You'll note that in every case within this range a+b=9

Neat trick, Bob!

Gene

PS

Another neat trick is to take any two double digit non-equal numbers with "opposite" ones and tens values between ten and one hundred and subtract them from one another... the result will always be a multiple of nine.

A number in decimal notation between zero and one hundred can represented as a sum of the first digit times ten and the second number times one.

For example, 56 is 5*10+6*1 and 65 would be 6*10+5*1.

Let any arbitrary number from zero to one hundred be expressed as 10*a+1*b and it's "opposite" as 10*b+1*a....

then to subtract them means taking 10*a+1*b minus 10*b+1*a=

10a+b-10b-a=(10a-a)+(-10b+b)=9a-9b=9(a-b)

In other words the final result is the difference of the first and last digits times nine showing that the final result must be a multiple of nine..

To prove this you must go the other way, to show a necessary and sufficient condition..

Take the difference of any two non-equal digits and multiply them times nine and you can create an equation that shows that their difference has a peculiar result...

For any a, b, not equal and less than 10 and greater than zero.

9(a-b)

9a-9b

9a=10a-a and 9b=10b-b

9a-9b=(10a-a)-(10b-b)=10a+b - (10b+a) => two non equal numbers such that their "opposites" form a difference, ie ab and ba.

"ab" minus "ba", which shows our original state and thus shows that the relationship is necessary and sufficient.

QED.

ddongbap
12-20-2008, 07:30 AM
Bob, of all people to put down a chain letter... really? lol.

Rain
12-20-2008, 10:53 AM
Great. Just the right kind of mathematical fix I need for today. LOL