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Math Help - finding numbers - problem solving !

  1. #1
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    Smile finding numbers - problem solving !

    find the largest 7-digit number which satisfy these 2 properties 0-9
    - no two digits in the number are the same
    - each of the numberís digits divides into the number exactly

    a. FIND 3 digits which can't be in this number and EXPLAIN why we cannot use them
    b. what is the highest 7 digit number you can make? EXPLAIN why it is the largest 7 digit you can make
    c. what is the SMALLEST number you can make which ALSO satisfy the properties

    any answers, hints, help discussion etc greatly appreciated!! i'm very stuck on this and don't know how to work it out.
    please also try to keep it simple so i can understand thank you.
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  2. #2
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    Re: Finding numbers Ė problem solving!

    Let us try this line of attack. In order to make the number as large as possible, we'll use the digits 9,8,7, which we will put at the beginning of the number. Then the number must be divisible by \mathrm{lcm}(9,8,7)=504. The largest multiple of 504 not greater than 9876543 is 9876384. That has a repeated digit and so we reject that. We then keep subtracting 504 from our number until we get one that fits.

    And the answer is:
    Spoiler:
    9867312
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  3. #3
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    Re: finding numbers - problem solving !

    Hello, ilikepizza!

    I don't have this solved, but I have some ideas.


    Find the largest 7-digit number (0-9) which satisfy these 2 properties:
    . . no two digits in the number are the same
    . . each of the numberís digits divides into the number exactly

    a. Find 3 digits which can't be in this number and explain why we cannot use them.

    b. What is the highest 7 digit number you can make?

    c. What is the smallest number you can make?

    Let N \;=\;\_\,\_\,\_\,\_\,\_\,\_\,\_

    The zero (0) cannot be in N; division by zero is not allowed.

    The five (5) cannot be in N.
    If N is divisible by 5, it must end in 5. .(It can't end in zero.)
    . . We have an odd number: . \_\,\_\,\_\,\_\,\_\,\_\,5
    We select 6 more digits from {1,2,3,4,6,7,8,9} . . . and will include at least one even digit.
    And no even digit divides into an odd number.

    I'm not sure what the third digit is.

    I do know this:
    . . If N contains a 9, the sum of the digits must be a multiple of 9.
    . . Then we have: {1,2,3,4,6,7,8,9} which has a sum of 40.
    . . For the sum to be a multiple of 9, we must discard the 4.
    So we have: {1,2,3,6,7,8,9}

    To be divisible by 8, the rightmost 3-digit number must be divisible by 8.
    . . For example: . N \;=\; \_\,\_\,\_\,\_\, 2\,1\,6
    Then we must arrange {3,7,8,9} so that N is divisible by 7.

    I found one solution: . N\;=\;8\,9\,7\,3\,2\,1\,6
    . . It could be the largest, but I have no proof.


    Edit: I like your solution, Sylvia!
    Last edited by Soroban; April 24th 2012 at 07:34 AM.
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    Re: finding numbers - problem solving !

    thank you both.
    I understand that 0 cannot be used as it is not possible to divide by 0, but i'm having trouble understanding the reasoning for 5 (I don't really get it.)
    Could you possibly explain it in really simple terms? Also I think the 3rd number is 4 just not sure how to explain it.
    any ideas for what the smallest number could be?

    Edit: i understand now after reading over thanks!

    but still not sure about the smallest number
    Last edited by ilikepizza; April 25th 2012 at 02:37 AM.
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  5. #5
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    Re: finding numbers - problem solving !

    Quote Originally Posted by ilikepizza View Post
    Also I think the 3rd number is 4, just not sure how to explain it.
    Any ideas for what the smallest number could be?
    The third number is 4 but only if you want to use the digit 9. We already know that 0 and 5 cannot occur in our number; let's show that 4 and 9 cannot both occur in our number. Suppose they do. A number is divisible by 9 if and only if the sum of its digits is divisible by 9 so let's check for divisibility by 9 for the candidates for our number:

    9 + 8 + 7 + 6 + 4 + 3 + 2 = 39

    9 + 8 + 7 + 6 + 4 + 3 + 1 = 38

    9 + 8 + 7 + 6 + 4 + 2 + 1 = 37

    9 + 8 + 7 + 4 + 3 + 2 + 1 = 34

    9 + 8 + 6 + 4 + 3 + 2 + 1 = 33

    9 + 7 + 6 + 4 + 3 + 2 + 1 = 32

    None of these is divisible by 9. Hence if we use the 4 we can't use the 9 Ė but we do want to use the 9 so as to make our number as large as possible.
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    Re: finding numbers - problem solving !

    yes, I get it now, thanks!
    is the smallest number 1289736? that was what I got
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  7. #7
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    Re: finding numbers - problem solving !

    Quote Originally Posted by ilikepizza View Post
    is the smallest number 1289736? that was what I got
    Yes.
    There are 105 total solutions:
    1: 1289736
    2: 1293768
    ...
    104: 9812376
    105: 9867312

    Little "looper" that'll get 'em all:
    LOOP a,b,c,d,e,f,g from 1 to 9
    DOWHILE variables all different and <> 4 or 5
    n = a*10^6 + b*10^5 + c*10^4 + d*10^3 + e*100 + f*10 + g
    PRINT n if n divisible bt all variables

    Have you "seen the light" on the "5" yet?
    To be divisible by 5, n must end with 5.
    If n ends with 5, then not divisible by even digits.
    Capish?
    Last edited by Wilmer; April 25th 2012 at 05:36 AM.
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