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

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Barry Schwarz

Posts: 80
Registered: 3/13/08
Re: Clever curiiosity --prove?
Posted: Aug 26, 2013 4:52 PM
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On Mon, 26 Aug 2013 12:57:08 -0500, Pfsszxt@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.


There are ten possible sets of numbers {a, b, c} that can be chosen
where the c is 30 or less. It is trivial to show that for these ten
cases, the assertion is true.

For any sequence where the final number exceeds 30, the number are of
the form {k*30+a, k*30+b, k*30+c} where a, b, and c form one of the
previous sets with c <= 30. The sum of these digits is obviously
3*k*30 + a+b+c
which will resolve to k*90 + 6. This can be expressed as
(k-1) * 90 +90 +6 =>
(k-1)*90 + 9 + 6 =>
(k-1)*90 +15 =>
(k-1)*90 + 6
Repeating this process another k-1 times produces a final sum of 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.


Each assertion has only 10 distinct cases. Exhaustive computation
seems the easiest way.

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