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Thread: divisible

  1. #1
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    divisible

    Prove that

    $\displaystyle 1^{1987} + 2^{1987} + ... + 13^{1987}$

    is divisible by $\displaystyle 15$.
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  2. #2
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    Hello,
    Quote Originally Posted by perash View Post
    Prove that

    $\displaystyle N=1^{1987} + 2^{1987} + ... + 13^{1987}{\color{red}+14^{1987}}$

    is divisible by $\displaystyle 15$.
    I guess there is a typo.. (in red)
    $\displaystyle 14 \equiv -1 (\bmod 15)$
    $\displaystyle 13 \equiv -2 (\bmod 15)$
    $\displaystyle 12 \equiv -3 (\bmod 15)$
    $\displaystyle 11 \equiv -4 (\bmod 15)$
    etc...

    So $\displaystyle N=1^{1987} + 2^{1987} + ... + 13^{1987}+14^{1987} \equiv 1^{1987} + 2^{1987}+3^{1987}$$\displaystyle +4^{1987}+5^{1987}+6^{1987}+7^{1987}-7^{1987}-6^{1987}-5^{1987}-4^{1987}-3^{1987}-2^{1987} -1^{1987} (\bmod 15)$

    $\displaystyle N \equiv 0 (\bmod 15)$
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