 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Ninety-Nine Nuance (Posted on 2014-06-11) Find the minimum positive integer N containing each of the digits from 1 to 9 exactly once such that N is divisible by 99.

 No Solution Yet Submitted by K Sengupta Rating: 3.0000 (1 votes) Comments: ( Back to comment list | You must be logged in to post comments.) Analytical Solution (spoiler?) | Comment 1 of 3
Well, the sum of the digits = 45, so it is guaranteed to be divisible by 9.  All we need is that it is also divisible by 11, and this occurs if the sum of the odd digits - sum even digits is divisible by 11.  One of them must be 28 and one must be 17.

Let's examine the minimum possible N, 123456789.
The sum of the odd digits = 25.  Sum of the even digits = 20.
To minimize N, play with the larger (rightmost) digits.  We need to increase the sum of the odd digits by 3, and this cannot be done without touching the 5.  Swap the 8 and the 5.  The new odd digits are now 1 + 3 + 7 + 8 + 9 = 28
The new even digits are now 2 + 4 + 5 + 6 = 17.

Minimum N, I think, is 123475869.

Edited on June 11, 2014, 1:41 pm
 Posted by Steve Herman on 2014-06-11 13:41:05 Please log in:

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