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Thread: 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 $\displaystyle n$ is not divisible by $\displaystyle 7$. The remainder when $\displaystyle n^2$ is divided by 7 and the remainder when $\displaystyle n$ is divided by $\displaystyle 7$ are each equal to $\displaystyle k$. What is $\displaystyle k$?

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

    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:
    $\displaystyle \[n = 7m + k\]$
    Therefore $\displaystyle \[{n^2} = 49 \cdot {m^2} + 14 \cdot m \cdot k + {k^2} = 7 \cdot (7 \cdot {m^2} + 2 \cdot m \cdot k) + {k^2}\]$

    We actually have to keep in mind, that we are trying to evaluate $\displaystyle \[k\]$ itself and only remainder of $\displaystyle \[{k^2}\]$ when it's divided by 7.
    However, in your case there really are no other possible scenarios other than $\displaystyle \[k = {k^2} = 1\]$ (you can run through $\displaystyle \[k = \overline {1,6} \]$ to make sure of that).
    Last edited by Pranas; Jan 14th 2011 at 06:40 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 $\displaystyle n$ is not divisible by $\displaystyle 7$. The remainder when $\displaystyle n^2$ is divided by 7 and the remainder when $\displaystyle n$ is divided by $\displaystyle 7$ are each equal to $\displaystyle k$. What is $\displaystyle k$?

    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.
    $\displaystyle \displaystyle\frac{n^2}{7}=m+\frac{k}{7}\Rightarro w\ n^2-7m-k=0$

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

    Substitute k

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

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


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

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

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

    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
    $\displaystyle \displaystyle\frac{n^2}{7}=m+\frac{k}{7}\Rightarro w\ n^2-7m-k=0$

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

    Substitute k

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

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


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

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

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

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

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

    If you concluded, that there is only one appropriate value for $\displaystyle \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 \displaystyle \((n+7)(n+x-m)=n^2+(-1)n+7(x-m)\Rightarrow\ x-m=-8\Rightarrow\ m-x=8$

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

    If you concluded, that there is only one appropriate value for $\displaystyle \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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