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

To find a nine-digit number that satisfies the given conditions, we need to approach this problem step by step. Let's break down the conditions and determine the possible options for each digit:

1. The first digit is divisible by 1.
This condition is satisfied by any number from 1 to 9.

2. The first two digits are divisible by 2.
In order for the first two digits to be divisible by 2, the last digit (unit's place) must be an even number. The possibilities are: 2, 4, 6, and 8.

3. The first three digits are divisible by 3.
To ensure that the sum of the first three digits is divisible by 3, we need to choose numbers that, when summed, have a remainder of 0 when divided by 3.
The possibilities for the last digit (unit's place) are 2, 4, 6, and 8, and the possibilities for the first two digits are:
- If the last digit is 2 or 8: 1 and 5.
- If the last digit is 4: 1 and 7.
- If the last digit is 6: 3 and 9.

4. The first four digits are divisible by 4.
To ensure that the last two digits are divisible by 4, we need to find pairs of numbers, where the last two digits can be rearranged to form a number divisible by 4.
The possible pairs are:
- If the last digit is 2: 12, 16, 32, 36, 52, 56, 72, 76, 92, and 96.
- If the last digit is 4: 24, 28, 64, 68, and 84.
- If the last digit is 6: No pairs are possible.
- If the last digit is 8: 48, 72, and 96.

Continuing this pattern, we can check all possible combinations and find the nine-digit number that satisfies all conditions.

Let's start by considering the previous example you provided, 923,156,784, and check if it meets the given conditions:
- The first digit, 9, is divisible by 1 (Condition satisfied).
- The first two digits, 92, are divisible by 2 (Condition satisfied).
- However, the first three digits, 923, are not divisible by 3 (Condition not satisfied).

Therefore, we need to find another nine-digit number that fulfills all the conditions.