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

 Messages: [ Previous | Next ]
 kiru.sengal@gmail.com Posts: 29 Registered: 12/7/05
Re: Clever curiiosity --prove?
Posted: Aug 26, 2013 2:24 PM

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.
>
>
> If the sum as multiple digits add the digits. Else done.
>
> Again if multiple digits, add them, etc etc.
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> Once you arrive at a single digit number the result will be 6.
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>
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> If the three consecutive numbers had the smallest divisible by three,
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> the result will always be 3.
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> If the middle one , etc, thre result will always be 9.
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>
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> 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

Date Subject Author
8/26/13 Pfsszxt@aol.com
8/26/13 kiru.sengal@gmail.com
8/26/13 Barry Schwarz
8/26/13 Ben Bacarisse
8/26/13 William Elliot