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