When n = 1, n must divide the first digit, therefore the first digit can only be 1

The second digit can be 2, 4, 6, or 8, since all of these numbers are divisible by 2


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


Of the numbers remaining to be used, the fourth digit can only be 6

Similar to n=1, when n=5, the fifth digit must be either 5 or 0; zero is not an option

There are only four numbers left, 4, 7, 8 and 9


There are only 3 numbers left, 7, 8 and 9, but none of then seem to work



We’ll begin again










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.