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

Home > Just Math
39 does it (Posted on 2012-02-19) Difficulty: 2 of 5
Prove that within any 39 consecutive natural numbers there must exist at least one number with the sum of its digits divisible by 11.
Bonus: Can the bound (39) be lowered?

See The Solution Submitted by Ady TZIDON    
Rating: 5.0000 (3 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution solution I think--based on computer exploration | Comment 2 of 4 |
(In reply to computer exploration -- bonus only is shown by Charlie)

Taking away the strict inequality and showing subsequent occurrences of the maximum gap shows:

38999980 to 39000019 is another such pair, as are 47999980 to 48000019, etc. where the first two digits add to 11 and 12 respectively.

In fact, come to think of it, doesn't this constitute proof? Whatever the sum of the digits in the 10 million and million positions is, or the congruence of any higher order digits mod 11, the cycle of 1 million in that range has only one occurrence that requires the 39 and none that require more. Quite a few require 29, which seems to be the second largest requirement--that is 28 is the second largest gap found in each set of 1 million numbers.

  Posted by Charlie on 2012-02-19 12:49:08
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 (1)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

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