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Thread: sum of the numbers less than p divisible by p?

  1. #1
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    sum of the numbers less than p divisible by p?

    is $\displaystyle \sum_{n=1}^{p-1}{n}$ always divisible by p if p is prime?

    Thanks for any help
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  2. #2
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    Quote Originally Posted by hmmmm View Post
    is $\displaystyle \sum_{n=1}^{p-1}{n}$ always divisible by p if p is prime?

    Thanks for any help
    Yes, unless $\displaystyle p=2$. In fact, it is true as long as $\displaystyle p$ is ANY odd number.

    Try rewriting the sum (for $\displaystyle p>2$, in which case $\displaystyle p$ is odd):

    $\displaystyle [1+(p-1)]+[2+(p-2)]+\cdots +[\frac{p-1}{2}+\frac{p+1}{2}]$
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  3. #3
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    $\displaystyle \sum_{n=1}^{p-1}{n}=\frac{p-1}{2}*p$ as p-1 is even (so divisible by 2) then it is divisible by p?
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  4. #4
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    Exactly. Writing the sum as I did above, you immediately get what you have just written.
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