N is a positive integer and S(N) denotes the sum of the digits.
Find all values of a positive integer constant C such that:
S(N) = C*S(N+3)
If N ends in 0,1,2,3,4,5,6 then S(N)<S(N+3) so C<1 and there is no solution.
If N ends in 7,8,9 then S(N)-S(N+3)=6 except
if there are more carries when 3 is added.
If there are a total of A carries then S(N)-S(N+3)=9A-3.
Now substitute the original equation for S(N) to get
C*S(N+3) - S(N+3) = 9A-3
So to achieve any C it suffices to find 9A-3 that is divisible by C-1. But since 9A-3 is never divisible by 9, C-1 cannot be a multiple of 9. So C is never 1 more than a multiple of 9.
There is still a presumption that there is a value of A that makes 9A-3 divisible by any other value of C-1. To see this is true just factor into 3(3A-1) and make (3A-1)=C-1 or 2(C-1) or 3(C-1).
Posted by Jer
on 2016-05-28 19:20:35