This has finally put me over the edge!!!!

[Teresa]

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??...



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

[Jo McDonald]

She missed me on both times I tried it..........
Boo Boo Hiss Hiss  Bad game.

[Patriot]

Come with me, little girl, on a magic carpet ride!  

Here's your answer...   http://wiki.answers.com/Q/How_does_Regifting_Robin_work

[Janet Harrington]

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.

[Patriot]

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.

[Jane]

 ::)She only guessed one out of my 5.

[Warph]

Come with me, little girl, on a magic carpet ride!  

Here's your answer...   http://wiki.answers.com/Q/How_does_Regifting_Robin_work




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.

[Varmit]

nope...she missed.