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

Quote:

---

Originally Posted by hmmmm

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):

$\displaystyle [1+(p-1)]+[2+(p-2)]+\cdots +[\frac{p-1}{2}+\frac{p+1}{2}]$

$\displaystyle \sum_{n=1}^{p-1}{n}=\frac{p-1}{2}*p$ as p-1 is even (so divisible by 2) then it is divisible by p?

Exactly. Writing the sum as I did above, you immediately get what you have just written.