Philippa M. Rhodes
A nine-digit number is formed using each of the digits 1,2,3,...,9 exactly once. For n = 1,2,3,...,9, n divides the first n digits of the number. Find the number.
There are several things to consider before using trial and error to find the number. First, we know that the fifth digit has to be a 5 so that the five-digit number is divisible by 5.
Next, we know that the even spaces have to be filled with even digits. We can even go a step further to say that either 2 or 6 will be in the 4th or 8th position leaving 4 or 8 to go in the 2nd and 6th positions. The divisibility rule for 4 is that the last two digits of a number must be divisible by 4. Since we know that the number in the 3rd position is an odd number, we can look at the numbers that are divisible by 4 in the teens, 30's, 50's, 70' and 90's and notice a pattern. All of these numbers end with a 2 or a 6.
The divisibilty rule for 8 is that the last 3 digits of a number must be divisible by 8. Similarly, the numbers in the 200's, 400's, 600's, and 800's that are divisible by 8 follow a pattern where the numbers that are in the teens, 30's, 50's, 70' and 90's of these hundreds end with either a 2 or a 6.
Then, after some trial and error, I found that 381654729 is a solution.