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Re: Clever curiiosity prove?
Posted:
Aug 26, 2013 4:52 PM


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 (k1) * 90 +90 +6 => (k1)*90 + 9 + 6 => (k1)*90 +15 => (k1)*90 + 6 Repeating this process another k1 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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