This should be in Misc.. but to be honest.. it is driving me so crazy that I have to post it here first.. ( I will move it later)
How the??.. What the??...
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This will drive you crazy............or it better.. cause I don't want to be in the loony bin all by myself.. ???
http://www.regiftable.com/regiftingrobinpopup.html
She missed me on both times I tried it..........
Boo Boo Hiss Hiss Bad game.
Come with me, little girl, on a magic carpet ride! ;D
Here's your answer... http://wiki.answers.com/Q/How_does_Regifting_Robin_work
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Well, she did not get me. I used the number 10. 1 - 0 = 1. 1 was a cell phone and she guessed a mouse pad. Ha ha.
Quote from: Janet Harrington on January 08, 2011, 03:24:43 PM
Well, she did not get me. I used the number 10. 1 - 0 = 1. 1 was a cell phone and she guessed a mouse pad. Ha ha.
I think you failed to follow the instructions..... If you chose 10, then the math would be: 10 - 0 - 1 = 9
Try again.
::)She only guessed one out of my 5.
Quote from: Patriot on January 08, 2011, 12:23:29 PM
Come with me, little girl, on a magic carpet ride! ;D
Here's your answer... http://wiki.answers.com/Q/How_does_Regifting_Robin_work
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Every time you subtract the first and second digits of the 2-digit number you chose, it comes up with a multiple of 9.
For example, 25-2-5=18. 18 is a multiple of 9.
So is: 75-5-7=63. 63 is a multiple of 9.
On the Regifting Robin gift list, all of the numbers that are a multiple of 9 have the same gift.
For example, let's say you did: 99-9-9=81.
The number you got will always be a multiple of 9, so all the multiples of 9 get the same gift:
So,
18=labelmaker
27=labelmaker
36=labelmaker
45=labelmaker
and so on and so forth. The gift changes every time to make it more convincing. Labelmaker is an example of a gift. It won't be labelmaker every time.
So the question should be why is it that we always end up with multiples of 9?
Let the number N = Ones+ 10x + 100y + 1000z + ... (example : 574 = 4 + 7*10 + 5*100 )
Its digits are w, x, y, z, ... So the sum of the digits is w + x + y + z + ... = S.
So, N - S = D = (ones-ones) + (10x-x) + (100y-y) + (1000z -z) + ... = 9x + 99y + 999z + ... = 9 (x + 11y + 111z + ...)
So, N - S is a multiple of 9.
But, there is more:
Once again let N = ones + 10x + 100y + 1000z + ...
We want to prove that, if the sum of its digits, S is divisible by 9, then so is the number, N, and conversely.
If S is divisible by 9, then S = 9p. Now add the number D to this, calculated above. S + D = 9p + D. But we saw previously that D itself is a multiple of 9, so D = 9q. Hence, S + D = 9p + 9q = 9(p+q) = 9r, say. But then, S + D is nothing but N. So N = 9r, which is what we wnted to show?
The converse is proved by starting with N = 9r and subtracting D to yield S = 9(r-q) = 9p,say.
Read more: http://wiki.answers.com/Q/How_does_Regifting_Robin_work#ixzz1Fc9ZC4RP
aaaah, yeah... I guess.
nope...she missed.