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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.

No Solution Yet Submitted by K Sengupta    
Rating: 3.0000 (1 votes)

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Solution Possible solution | Comment 2 of 3 |

Any number that contains each of the digits from 1 to 9 exactly once will be divisible by 9.  So the problem is to find N such that N is divisible by 11.

The test for divisibility by 11 is to sum every second digit and subtract the sum of all the other digits; if the difference is 0 mod 11 then so is N.  Note that other than this grouping, order doesn't matter.

When N contains each digit from 1 to 9 exactly once, the difference between these sums will always be odd.  So we're looking to group the digits such that the difference of their sums is + or - 11, so the sums will be 17 and 28. 

We can make 17 with (2,4,5,6) so ordering everything from least to greatest would make N = 123475869 = 1247231 * 99.

It seems to me that would be the minimum solution.


  Posted by tomarken on 2014-06-11 14:03:44
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