From: i.still@hate.spam.be   
      
   Sampath wrote:   
      
   > Sum of the digits in a number :   
   >   
   > Consider 7291   
   >   
   > Sum of the digits,   
   >   
   > 7 + 2 + 9 + 1 = 19 = 1 + 9 = 10 = 1 + 0 = 1   
   >   
   > 72 + 91 = 163 = 16 + 3 = 19 = 10 = 1   
   >   
   > 79 + 21 = 100 = 1   
   >   
   > 27 + 91 = 118 = 11 + 8 = 19 = 1   
   >   
   > ...   
   >   
   > You will get the same sum from the digits.   
   >   
   > Can you form a statement for this ?   
      
   Easy: your result always holds.   
      
   Denote e.g. a four-digit number by a/b/c/d (for instance the number   
   7291 is denoted as 7/2/9/1).  Choose a random ordering of the digits   
   and a random splitting in two or more different numbers, e.g. b/c and   
   d/a.  Then b/c + d/a = (b+d)/(c+a).  If c+a < 10, then we obtain that   
   (b+d) + (c+a) = a+b+c+d, so the sum of the digits of the newly   
   created number (b+d)/(c+a) is the same as the sum of the digits of   
   the original number.  If c+a >= 10, then (b+d)/(c+a) = e/f, with f =   
   c+a-10 and e = b+d+1.  Hence the sum of the digits of e/f is equal to   
   e+f = (b+d+1) + (c+a-1) = a+b+c+d (we have c+a-1 because if c+a >=   
   10, then c+a-10 is a one-digit number which misses the digit 1 from   
   c+a).  So the sum of the digits of the new number is also the same as   
   the sum of the digits of the original number.  Of course, the   
   reasoning in the other cases is similar.   
      
   Probably this is proven already in a book on number theory (I'm too   
   lazy to check this out).   
      
   --   
   If I have seen further it is by standing on ye shoulders of   
   Giants -- Isaac Newton.   
      
   --- SoupGate-Win32 v1.05   
    * Origin: you cannot sedate... all the things you hate (1:229/2)