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)=9A3.
Now substitute the original equation for S(N) to get
C*S(N+3)  S(N+3) = 9A3
S(N+3)=(9A3)/(C1)
So to achieve any C it suffices to find 9A3 that is divisible by C1. But since 9A3 is never divisible by 9, C1 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 9A3 divisible by any other value of C1. To see this is true just factor into 3(3A1) and make (3A1)=C1 or 2(C1) or 3(C1).

Posted by Jer
on 20160528 19:20:35 