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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. ??

  • math -

    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

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