All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Just Math
Digit Sum Equation Puzzle (Posted on 2016-05-28) Difficulty: 3 of 5
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)

No Solution Yet Submitted by K Sengupta    
No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution A more full solution. | Comment 4 of 5 |
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 addedIf 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
Please log in:
Remember me:
Sign up! | Forgot password

Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (5)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Copyright © 2002 - 2018 by Animus Pactum Consulting. All rights reserved. Privacy Information