When n = 1, n must divide the first digit, therefore the first digit can only be 1
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The second digit can be 2, 4, 6, or 8, since all of these numbers are divisible by 2
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Recall casting out 3’s,
If x is the third digit, then 1 + 2 + x is a multiple of 3
Using this rule, only 3, 6, or 9 can be used
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Of the numbers remaining to be used, the fourth digit can only be 6
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Similar to n=1, when n=5, the fifth digit must be either 5 or 0; zero is not an option
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There are only four numbers left, 4, 7, 8 and 9
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There are only 3 numbers left, 7, 8 and 9, but none of then seem to work
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We’ll begin again
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Notice that no matter what the order of the digits are, the sum is always 45.
This means that a nine digit number using the numbers 1 through 9 will always be divisible by 9.
Also notice, the numbers are in sequential order, with the exception of 2 and 8.