math
posted by maddi 5 on .
Using the numbers 1 through 9 with no repeats, find a 9 number such that: the first digit is divisible by 1, the first two digits are divisible by 2, the first 3 digits are divisible by 3, and so on until we get to a 9 digit number divisible by 9. You might try, for example, the number 923,156,784. But this number doesn't work the first three digit number, 923, is not divisible by 3. Find the nine digit number that works. ??

I am curious at what level of math this is from.
It looks like you are looking at divisibility of numbers
Here are some basic rules:
 any number of course is divisible by 1
 to be divisible by 2 , the number has to be even
so the 2nd digit must be even
 to be divisible by 3, the sum of the digits of the number has to be divisible by 3
 to be divisible by 4, the last 2 digits must be divisible by 4 , and the last digit must of course be even
 to be divisible by 5, the last digit must be either 0 or 5, but we don't have a 0, so the 5th digit MUST be 5, ahhh, that's a start
 to be divisible by 6, it must be divisible by 2 AND by 3, that is, it must be even and pass the divisible by 3 test.
 to be divisible by 7 ,,,, there is a way, but much too complicated to state in simple terms
 to be divisible by 8, the last 3 digits must be divisible by 8, and of course the last digit must be even
 to be divisible by 9, the sum of the digits must be divisible by 9, which in this case is true no matter how they are arranged.
so far we know:
the 5th digit must be 5
the 2nd, the 4th, the 6th, and the 8th digits must be even
e.g. The 1 cannot be in the place it is in
I suggest you cut out little pieces of paper, label them 1, 2, ... 9 and try it.
Let me know what number you come up with