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Math Help - Positive integer not divisible by 7

  1. #1
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    Positive integer not divisible by 7

    Hi,


    Is there a simple explanation for this?

    The positive integer n is not divisible by 7. The remainder when n^2 is divided by 7 and the remainder when n is divided by 7 are each equal to k. What is k?

    A. 1
    B. 2
    C. 4
    D. 6
    E. It cannot be determined from the information given.

    I just looked at the middle part and concluded the answer to be 1. The only numbers I know that are equal to their squares are 0 and 1. Zero isn't available.
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  2. #2
    MHF Contributor Unknown008's Avatar
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    I can't see any answer besides A. ..

    Note that it cannot be zero because otherwise, n would be a multiple of 7.
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  3. #3
    Member Pranas's Avatar
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    Quote Originally Posted by Hellbent View Post
    Hi,


    Is there a simple explanation for this?

    The positive integer n is not divisible by 7. The remainder when n^2 is divided by 7 and the remainder when n is divided by 7 are each equal to k. What is k?

    A. 1
    B. 2
    C. 4
    D. 6
    E. It cannot be determined from the information given.

    I just looked at the middle part and concluded the answer to be 1. The only numbers I know that are equal to their squares are 0 and 1. Zero isn't available.
    I have to agree.
    Let's say our number can be written as:
    \[n = 7m + k\]
    Therefore \[{n^2} = 49 \cdot {m^2} + 14 \cdot m \cdot k + {k^2} = 7 \cdot (7 \cdot {m^2} + 2 \cdot m \cdot k) + {k^2}\]

    We actually have to keep in mind, that we are trying to evaluate \[k\] itself and only remainder of \[{k^2}\] when it's divided by 7.
    However, in your case there really are no other possible scenarios other than \[k = {k^2} = 1\] (you can run through \[k = \overline {1,6} \] to make sure of that).
    Last edited by Pranas; January 14th 2011 at 07:40 AM.
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  4. #4
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    Quote Originally Posted by Hellbent View Post
    Hi,


    Is there a simple explanation for this?

    The positive integer n is not divisible by 7. The remainder when n^2 is divided by 7 and the remainder when n is divided by 7 are each equal to k. What is k?

    A. 1
    B. 2
    C. 4
    D. 6
    E. It cannot be determined from the information given.

    I just looked at the middle part and concluded the answer to be 1. The only numbers I know that are equal to their squares are 0 and 1. Zero isn't available.
    \displaystyle\frac{n^2}{7}=m+\frac{k}{7}\Rightarro  w\ n^2-7m-k=0

    \displaystyle\frac{n}{7}=x+\frac{k}{7}\Rightarrow\ n-7x=k

    Substitute k

    n^2-7m-n+7x=0\Rightarrow\ n^2-n+7(x-m)=0

    \Rightarrow\ (n+7)(n+x-m)=n^2+(-1)n+7(x-m)\Rightarrow\ x-m=-8\Rightarrow\ m-x=8


    (n-8)(n+7)=0\Rightarrow\ n=8

    \displaystyle\frac{n^2}{7}=\frac{64}{7}=9\frac{1}{  7}

    \displaystyle\frac{8}{7}=1\frac{1}{7}

    The remainder is 1.
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  5. #5
    Member Pranas's Avatar
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    Quote Originally Posted by Archie Meade View Post
    \displaystyle\frac{n^2}{7}=m+\frac{k}{7}\Rightarro  w\ n^2-7m-k=0

    \displaystyle\frac{n}{7}=x+\frac{k}{7}\Rightarrow\ n-7x=k

    Substitute k

    n^2-7m-n+7x=0\Rightarrow\ n^2-n+7(x-m)=0

    \Rightarrow\ (n+7)(n+x-m)=n^2+(-1)n+7(x-m)\Rightarrow\ x-m=-8\Rightarrow\ m-x=8


    (n-8)(n+7)=0\Rightarrow\ n=8

    \displaystyle\frac{n^2}{7}=\frac{64}{7}=9\frac{1}{  7}

    \displaystyle\frac{8}{7}=1\frac{1}{7}

    The remainder is 1.
    Could you please clarify how did you come up with
    \displaystyle \((n+7)(n+x-m)=n^2+(-1)n+7(x-m)\Rightarrow\ x-m=-8\Rightarrow\ m-x=8

    and \displaystyle \[{\rm{n = 8}}\]

    If you concluded, that there is only one appropriate value for \displaystyle \[{n}}\] under given circumstances, that is not true.
    Otherwise my apologies if I misunderstood something.
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  6. #6
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    Quote Originally Posted by Pranas View Post
    Could you please clarify how did you come up with
    \displaystyle \((n+7)(n+x-m)=n^2+(-1)n+7(x-m)\Rightarrow\ x-m=-8\Rightarrow\ m-x=8

    and \displaystyle \[{\rm{n = 8}}\]

    If you concluded, that there is only one appropriate value for \displaystyle \[{n}}\] under given circumstances, that is not true.
    Otherwise my apologies if I misunderstood something.
    Of course, multiples of 7 may be added to n=8 for a full solution.
    My earlier post outlines a "beginning" for another analysis.
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