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Math Help - 10 digit number

  1. #1
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    10 digit number

    A 10 digit number is made up of only 5s and 0s. It's also divisible by 9. How many possibilities are there for the number?

    Working backwards, I cam e to the conclusion that it cannot end in a 0, as that would been there was a remainder of 9 from the previous number. Then if it ends in a 5, that means a remainder of 4 was there from the previous. Now that could mean the 2nd last digit is either 0 or 5. But I'm lost after that, don't think I can apply this reasoning till the first digit.
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  2. #2
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    So you are solving \sum_{n=0}^9 a_n10^n\equiv 0 (\bmod 9), where each a_n\in\{0,5\}. Since 10^n\equiv 1 (\bmod 9) for all n, all you have is 9|\sum_{n=0}^9 a_n. Since a_n is either 0 or 5, you really have 9|5k for some k, which can only be 0 or 9. So you either have all zeroes, no fives, or 9 fives, 1 zero. There is one solution for the first option, and _{10}P_1=10 solutions for the second option, so your answer is 11.
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  3. #3
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    Hi Media Man.

    Sorry I didn't follow what you did on:

    Quote Originally Posted by Media_Man View Post
    Since 10^n\equiv 1 (\bmod 9) for all n, all you have is 9|\sum_{n=0}^9 a_n. Since a_n is either 0 or 5, you really have 9|5k for some k, which can only be 0 or 9.
    So you divided the coefficients of each power of 10 by 9, and since each 10 leaves a remainder of 1, that equals 9 so they can be ignored. OK

    Then how do you the next bit with the 5k?
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  4. #4
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    The jump to "so 9|5k" just means you are looking for a 10 digit number consisting of k 5's in any order whatsoever. In other words, the power of 10 for which they are coefficients is irrelevant. So 505=550=055 (mod 9), for example.
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  5. #5
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    a_9 \cdot 10^9 + a_8 \cdot 10^8 + a_7 \cdot 10^7 + ... + a_3 \cdot 10^3 + a_2 \cdot 10^2 + a_1 \cdot 10^1 + a_0 \cdot 10^0 \equiv 0 \ (\bmod \ 9)

    Since 10^n\equiv 1 \ (\bmod \ 9)

    a_9 \cdot 1 + a_8 \cdot 1 + a_7 \cdot 1 + ... + a_3 \cdot 1 + a_2 \cdot 1 + a_1 \cdot 1 + a_0 \cdot 1 \equiv 0 \ (\bmod \ 9)

    Since the a's are either 0 or 5, when you add them, it's basically a multiple of 5, right? (if there are x a's that are equal to 5, then the sum of the digits will be 5x)

    That's how Media_Man got: 5k \equiv 0 \ (\bmod \ 9) i.e. 9|5k, where 0 \leq k \leq 10

    You can think of it this way, do you know how to tell if a number is divisible by 9? (The sum of the digits should be divisible by 9, we kind of have a modified proof here for a 10 digit number) So, how many 5's should the number have so that when you sum up all the 5's, it should be divisible by 9?

    I think the answer is actually 9 ... since 0 isn't a 10 digit number and 0555555555 isn't a 10 digit number either ... I guess it depends though ...

    You can also check those 9 numbers by dividing them by 9 ... it's only 9 numbers
    Last edited by Bingk; November 2nd 2009 at 01:42 PM. Reason: Added how to check
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  6. #6
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    There is a really simple way to solve this.
    If an integer is divisible by nine the sum of the digits is multiple of nine.
    To have a ten digit number then the string must begin with a 5 followed by eight 5ís and one 0 in any order.
    So the answer is 9.
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