posted by maddi 5 .
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