Take the set $\{1,2,3,4,5,6\}$ and any permutation of it. and then add the 1 to the first element of that permutation, 2 to the second member of the permutation and so on. Denote X as the set of the 6 sums.

Let $X=\{a_1,a_2,a_3,a_4,a_5,a_6\}$ can it be that $a_i\neq a_j \mod6$ for any $i\ne j$


No, because if the six elements of $X$ all have different residues mod 6, then these residues must be $0,1,2,3,4,5$ in some order, and so we would have $$\sum_{a_i\in X} a_i \equiv \sum_{i=0}^5 i = 15 \equiv 3\pmod 6.$$ But it should be clear that the sum of the elements of $X$ is actually $0\pmod 6$.

  • $\begingroup$ Might be worth pointing out that this argument only works when the original set $\{1,\ldots n\}$ has $n$ even. For $n$ odd, you can indeed have a permutation that goes the way you want. For example, take $n=5$ and $X = \{1+1, 2+2, 3+3, 4+4, 5+5\} = \{2, 4, 1, 3, 0\} \pmod 5$. $\endgroup$ – MJD Aug 5 '13 at 17:43
  • $\begingroup$ math.stackexchange.com/questions/460338/… $\endgroup$ – Jorge Fernández Hidalgo Aug 5 '13 at 18:03

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