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Math Help - remainder when divided by 7

  1. #1
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    remainder when divided by 7

    What is the remainder when (1^3)+(2^3)+(3^3)+...+(1990^3) is divided by 7?
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    MHF Contributor Also sprach Zarathustra's Avatar
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    Try to prove:
    If x is integer, then x^3(mod7)=0 or 1 or 6
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by chris86 View Post
    What is the remainder when (1^3)+(2^3)+(3^3)+...+(1990^3) is divided by 7?
    We need merely note that \displaystyle \left\lfloor\frac{1990}{7}\right\rfloor=284 and that \displaystyle 7\cdot 284=1988. From this we can note that

    \displaystyle \#\left\{m\leqslant 1990:m\equiv k\text{ mod }7\right\}=284\;\;\;k=0,\cdots,6


    It follows then that


    \displaystyle \sum_{n=1}^{1990}n^3\equiv 284\sum_{n=1}^{7}n^3+1989^3+1990^3\text{ mod }7


    Note though that in \mathbb{Z}/7\mathbb{Z} we have that 4^{-1}=2 and so in particular



    \displaystyle \sum_{n=1}^{1990}n^3\equiv 284\sum_{n=1}^{7}n^3+2\equiv 2\cdot284\cdot 4\sum_{n=1}^{7}n^3+2\text{ mod }7


    Note though that in general


    \displaystyle 4\sum_{k=1}^{m}k^3=m^2(m+1)^2


    And so in particular


    \displaystyle 2\cdot 284\cdot4\sum_{n=1}^{7}n^3+2\equiv 2\cdot284\cdot 7^2\cdot8^2+2\equiv 2\text{ mod }7
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    Quote Originally Posted by Drexel28 View Post
    We need merely note that \displaystyle \left\lfloor\frac{1990}{7}\right\rfloor=284 and that \displaystyle 7\cdot 284=1988. From this we can note that

    \displaystyle \#\left\{m\leqslant 1990:m\equiv k\text{ mod }7\right\}=284\;\;\;k=0,\cdots,6


    It follows then that


    \displaystyle \sum_{n=1}^{1990}n^3\equiv 284\sum_{n=1}^{7}n^3+1989^3+1990^3\text{ mod }7


    Note though that in \mathbb{Z}/7\mathbb{Z} we have that 4^{-1}=2 and so in particular



    \displaystyle \sum_{n=1}^{1990}n^3\equiv 284\sum_{n=1}^{7}n^3+2\equiv 2\cdot284\cdot 4\sum_{n=1}^{7}n^3+2\text{ mod }7


    Note though that in general


    \displaystyle 4\sum_{k=1}^{m}k^3=m^2(m+1)^2


    And so in particular


    \displaystyle 2\cdot 284\cdot4\sum_{n=1}^{7}n^3+2\equiv 2\cdot284\cdot 7^2\cdot8^2+2\equiv 2\text{ mod }7

    The above is too involved, imho. Using Zaratustra's hint, it's easy to show that

    (7k)^3+(7k+1)^3+\ldots +(7k+6)^3=0\!\!\pmod 7\,,\,\,\forall k\in\mathbb{N} , so since 1990=7\cdot 284+2 , we get

    that the sum equals  1989^3+1990^3=1^3+2^3=2\!\!\pmod 7

    Tonio
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