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Topic: Clever curiiosity --prove?
Replies: 4   Last Post: Aug 26, 2013 10:19 PM

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kiru.sengal@gmail.com

Posts: 29
Registered: 12/7/05
Re: Clever curiiosity --prove?
Posted: Aug 26, 2013 2:24 PM
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On Monday, August 26, 2013 1:57:08 PM UTC-4, Pfs...@aol.com wrote:
> Take ANY three consecutive numbers with the largest divisible by 3.
>
> Add them.
>
> If the sum as multiple digits add the digits. Else done.
>
> Again if multiple digits, add them, etc etc.
>
> Once you arrive at a single digit number the result will be 6.
>
>
>
> If the three consecutive numbers had the smallest divisible by three,
>
> the result will always be 3.
>
>
>
> If the middle one , etc, thre result will always be 9.
>
>
>
> Prove!!


First case:

Numbers are of form 3k+1, 3k+2, 3k+3 for some integer k

Sum: 9k+6

Thus remainder of this number when divided by 9 is 6.

Thus sum of it's digits divided by 9 is also 6 ( This is a separate proposition)

Proof of proposition:
-----------------------

The decimal expansion of an integer N is:

N = a_n*10^n + ... + a1*10^1 + a0*10^0 for some integers 0<=a_i<10

modulo 9, we have 10^i = 1^i =1
Thus,

N = a_n + .. + a1 + a0 (mod 9)

So the left side and right side have same remainder when divided by 9



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