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)
S(N+3) is lower than S(N) => last digit of N+3 =(0,1,2) and last digit of N = (7,8,9).
There is a trivial solution for all numbers 10^n - 3 because then S(N+3) = 1. For N+3=10^n, C=S(N)=7,16,25,....
If N is congruent with 1-96 (mod 100) then:
S(N)-S(N+3)=6 => (C-1)*S(N+3)=6
This is possible for values of S(N+3)=(6,3,2,1) and for the vlues of C= (2,3,4,7).
As the last digit of N+3 =(0,1,2) and S(N+3)=(1,2,3,6)
Combining all possibities;
N+3=(11,12,20,21,30,42,51,60,110,111,120,132,141,150,
210,222,231,240,312,321,330,411,420,510,600)
If N is congruent with 97-99 (mod 100) (C-1)*S(N+3)=15, so C=(4,6).
and there are values for N+3=(102,201,302,401,1200,
2100,1400,2300,3200,4100...
But probably there is a better way than this to solve this puzzle.
Edited on May 28, 2016, 12:18 pm
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Posted by armando
on 2016-05-28 11:33:44 |
No Subject
