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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:






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.

See the Related Link for more info on the workings of Regifting Robin.

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โˆ™ 2011-02-09 17:34:53
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