Question: "For which n does the expression 1+2+...+(n-1) congruent to 0(mod n) hold?"
If I am reading this correctly, for the expression 0 is the least positive residue of (1+2+...+(n-1))(mod n). However, if that is correct, I am not sure where to go from there. Would n be 0 or 1 then? Or something else entirely?

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Then if n=3, it would be 1+2=3 congruent to 0mod3, which would be true then, aye?

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