N is a positive integer entirely formed by the digits 0,1,2,3,5, and 7 such that N is evenly divisible by 297. (Each of these six digits must occur at least once in N. None of the remaining four digits can occur in N.)
Determine the minimum value of N.
First note 297 = 11 * 27 = 11*9*3
Let's assume there's a 6-digit solution, which therefore must be a permutation of the six permitted digits.
Since the sum of the allowed digits is 9, all of these permutations are divisible by 9.
Also, any number whose odd and even digits have the same sum is divisible by 11. Here, that means the odd digits must be (0,2,7) and the even (1,3,5) or vice versa. It then remains to check the various combinations for one that's not just a multiple of 9 but of 27.
Assuming there's no leading zeros permitted, the smallest such must start with 1 and ideally follow with 0, giving the 4 choices: 103257, 103752, 105237, and 105732. Dividing each by 99 (since we know they're all divisible by that much) gives 1043, 1048, 1063, and 1068. It's easy to see that the last of these, 1068, is divisible by 3 and the others aren't, so 105732 is divisible by 99*3 = 297.
Since there are no other solutions beginning with 10, and no solutions with fewer digits, this is the smallest solution.
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Posted by Paul
on 2024-07-30 12:33:18 |
Solution