please, know somebody solution for this argument?
Prove, that for each odd $m > 2$, it is true that $$\sum_{k=1}^{m-1} k \equiv 0 \pmod m$$
Thanks for yours answers!
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Hint: Try induction. – Sean Roberson May 03 '17 at 18:15
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1Where does $n$ appear in the equation? – kccu May 03 '17 at 18:15
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EDITOR: I'm not sure if this is what you intend. The original phrasing was "Prove, that for each odd modulus m > 2, is true that sum from k=1 to m-1 k ≡ 0 (mod m)". – Kenny Lau May 03 '17 at 18:15
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What is the modulus of the congruence, what role does $n$ play ? Please edit. – Rene Schipperus May 03 '17 at 18:16
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I think I edited this as the OP intended. I forgot to edit the title though. – John Doe May 03 '17 at 18:20
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$$\sum_{k=1}^{m-1} k = \frac{(m-1)m}{2}$$
Note that if $m$ is odd, then $m-1$ is an even integer, which means we can write $\frac{(m-1)m}{2} = am$ for some integer $a$. And $am \equiv 0 \bmod{m}$ by inspection.
Marcus Andrews
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@Dr.Zoidberg The $2$ is "absorbed" by the $m-1$ term. For instance, if $m=5$, we know $m-1 = 4$, and $4$ is divisible by $2$. So the $(m-1)/2$ part can be replaced by an integer $a$. – Marcus Andrews May 03 '17 at 18:31