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

Home > Just Math
Ninety-Nine Nuance (Posted on 2014-06-11) Difficulty: 3 of 5
Find the minimum positive integer N containing each of the digits from 1 to 9 exactly once such that N is divisible by 99.

See The Solution Submitted by K Sengupta    
Rating: 4.0000 (2 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
My path to the wrong answer--not a solution Comment 3 of 3 |
As N contains all the digits 1-9 exactly once each, it is necessarily divisible by 9, so all we need in addition is divisibility by 11.

123456789, the smallest such pandigital number, totals 25 in the odd positions and 20 in the even positions.  The difference is not a multiple of 11.  What's needed is either to get the difference down to zero or up to 11.

Interchanging two adjacent digits changes that odd-even difference by 2, so it's easier to get to a difference of 11, as 11 and 5 are both odd. Since we want to increase the difference by 6 we need to make three such interchanges, as near to the right side of the number as possible. As we want to increase the odd positions while decreasing the even sum, we swap 8 with 7, 6 with 5 and 4 with 3, making 124365879 the desired result.

  Posted by Charlie on 2014-06-11 14:55:07
Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (1)
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 - 2024 by Animus Pactum Consulting. All rights reserved. Privacy Information