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I am looking at the following problem: Given the finite set $\{2,3,4,5,6,7,8\}$ suppose we choose three distinct elements $a,b,c$ out of it. Clearly there are $\binom{7}{3}$ ways of doing that. Without going through all the cases I wish to prove that each case yields an $x,y\in\{a,b,c\}$ such that either $\{1,x,y\}$ contains no consecutive integers or $\{x,y,9\}$ contains no consecutive integers.

I have tried to cut down the cases by symmetric arguments, but can someone suggest a cleaner proof?


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up vote 2 down vote accepted

If $2 \notin \{a,b,c\}$, then $\{1,\min\{a,b,c\},\max\{a,b,c\}\}$ contains no consecutive integers.
If $8 \notin \{a,b,c\}$, then $\{\min\{a,b,c\},\max\{a,b,c\},9\}$ contains no consecutive integers.
And if $a=2$ and $b=8$, then either $\{1,c,8\}$ or $\{2,c,9\}$ contains no consecutive integers.

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Let the three chosen numbers be $a<b<c$. If $a=2$, then either $\{1,b,c\}$ works, or $c=b+1$. In that case $\{2,b,9\}$ works unless $b=3$, in which case $c=4$ and $\{2,4,9\}$ works. By symmetry we’re also okay if $c=8$, so assume that $3\le a<b<c\le 7$. Then $\{1,a,b\}$ works unless $b=a+1$, in which case $\{1,a,c\}$ works.

I don’t see any clever way to avoid all consideration of cases, but this seems pretty short.

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Order elements so that $a<b<c$. If $a,b,c$ are three consecutive numbers then $c-a=2$ and either $a\neq2$, in which case $\{1,a,c\}$ works, or $c\neq8$ in which case $\{a,c,9\}$ works. In the remaining case combine $x=b$ with some $y\in\{a,c\}\setminus\{b-1,b+1\}$ and then with some $z\in\{1,9\}\setminus\{y-1,y+1\}$ (both differences are nonempty, and adjacencies are avoided).

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