Find a pair of two consecutive integers, such that the s.o.d. of each of them is a multiple of 49. i.e. SOD(N)=K*49 & SOD(N+1)=L*49: (K,L integers)

Rem1: S.o.d. of a number is the sum of said number's digits.
Rem2: Only the smallest values requested.

One possible solution is to take a string of 9's placed after 499998 (sod is 48). This string of 9's, when 1 is added, will become 499999 (sod is 49), followed by a bunch of zeros. So now we need to find, how many 9's would be necessary for the original number to have a S.O.D. divisible by 49.

Since 499998 has a S.O.D. value of 48, we need our S.O.D. for 999.... to equal 1 mod 49. So, set 9x = 1 + 49k, and check for a few values of k:

k = 1, 1 + 49 = 50: 50/9 = 5.55555555

Edited on February 1, 2011, 3:20 pm

Posted by Justin on 2011-02-01 15:18:34