1 + 2 + 3 +...+ (p1)(mod p)
1. The problem statement, all variables and given/known data
What is the value of 1 + 2 + 3 +...+ (p1)(mod p)? 2. Relevant equations p = 0 (mod p) p1 = 1 (mod p) 1 + 2 + 3 + ...+n = n(n+1)/2 3. The attempt at a solution I know 1 + 2 + 3 +...+ (p1) = (p1)(p)/2 I worked the problem, but i don't know if i am correct: work: i am looking for a b s.t. b = (p1)(p)/2 (mod p) or 2b = (p1)(p) ( mod p) we know (p1) = 1 (mod p) so : (p1)p = p (mod p) this implies : 2b = p(mod p) ,also we know p = 0 (mod p) so b =0 (mod p) So the answer is 1 + 2 + 3 +...+ (p1) = 0 (mod p) 
Re: 1 + 2 + 3 +...+ (p1)(mod p)
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For example, if p = 6, 1 + 2 + 3 + 4 + 5 ##\equiv## 3 (mod 6). If p = 5, 1 + 2 + 3 + 4 ##\equiv## 0 (mod 5). 
Re: 1 + 2 + 3 +...+ (p1)(mod p)
BTW, what you're calculating is (1 + 2 + ... + (p 1))(mod p).

Re: 1 + 2 + 3 +...+ (p1)(mod p)
Mark, I am sorry I was not clear, when I say p I mean a prime number.

Re: 1 + 2 + 3 +...+ (p1)(mod p)
How about if p = 2, the only even prime?
1 ≠ 0 (mod 2) 
Re: 1 + 2 + 3 +...+ (p1)(mod p)
Well, you have the right idea by writting
[tex]1+2+\cdots+(p1) = \frac{(p1)p}{2}[/tex] But then you sort of take the long route to the eventual answer. When you have this: [tex]1 + \cdots + (p1) = \frac{(p1)p}{2}[/tex] just note that [itex]p1[/itex] is even so that [itex](p1)/2[/itex] is an integer. Thus, the sum is divisible by [itex]p[/itex] and so the answer is [itex]0[/itex]. But, yes, your answer is correct. EDIT: As Mark mentioned, you'll have to do a special case for 2. In fact, this happens a lot in Number Theory. 
Re: 1 + 2 + 3 +...+ (p1)(mod p)
I see.
So if p=2 then I know the sum is : 1 + 1 = 2, which is congruent to 0 mod 2 as well. Thank you both for you help! 
Re: 1 + 2 + 3 +...+ (p1)(mod p)
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