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

Home > Numbers
Forty- niners (Posted on 2011-02-01) Difficulty: 3 of 5
Find a pair of two consecutive integers, such that the s.o.d. of each of them is a multiple of 49. i.e. SOD(N)=K*49 & SOD(N+1)=L*49: (K,L integers)

Rem1: S.o.d. of a number is the sum of said number's digits.
Rem2: Only the smallest values requested.

No Solution Yet Submitted by Ady TZIDON    
Rating: 4.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
solution? | Comment 2 of 5 |
One possible solution is to take a string of 9's placed after 499998 (sod is 48). This string of 9's, when 1 is added, will become 499999 (sod is 49), followed by a bunch of zeros. So now we need to find, how many 9's would be necessary for the original number to have a S.O.D. divisible by 49.

Since 499998 has a S.O.D. value of 48, we need our S.O.D. for 999.... to equal 1 mod 49. So, set 9x = 1 + 49k, and check for a few values of k:

k = 1, 1 + 49 = 50: 50/9 = 5.55555555
k = 2, 1 + 49 * 2 = 99: 99/9 = 11

So, a string of eleven 9's, following 499998, or:

49,999,899,999,999,999

... seems like a good place to start.

Edited on February 1, 2011, 3:20 pm
  Posted by Justin on 2011-02-01 15:18:34

Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (6)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2017 by Animus Pactum Consulting. All rights reserved. Privacy Information