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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?

Thanks.

3 Answers 3

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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

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