Suppose the numbers $1,2,3,...,10$ are spilt into two disjoint collections $a_1,a_2,..a_5$ and $b_1,b_2,...,b_5$ such that

$a_1<a_2<a_3<a_4<a_5$ $b_1>b_2>b_3>b_4>b_5$

Show that the larger number in any pair ${a_j,b_j}, 1≤j≤5,$ is at least $6$.

It's a question from a Regional math contest. I don't understand how to approach it. The proposed solution (which I do not understand) is:

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Suppose $j=3$ then $a_3,b_3$ are two distinct numbers.

$a_1$ and $a_2$ are both less that $a_3$ so $a_3$ must be at least $3$

$b_4$ and $b_5$ are both less than $b_3$ so $b_3$ must be at least $3$

but we can't have $a_3=b_3$ so one of them must be at least $4$. That's what we get from the first attempt. But we can do better.

Now suppose $a_3$ and $b_3$ are both less than $6$ then $a_1, a_2, b_4, b_5$ (the four numbers referred to in the proof you have quoted) must all be less than $6$ too - that is six distinct positive integers less than $6$, which is impossible, so the largest, which will be $a_3$ or $b_3$, will be at least $6$.

The proof you have been given takes this observation and generalises it to any $j$ rather than just $3$. Always try to work out a specific example of a general statement if you don't quite understand what it is saying.

Another way of approaching this is to look at the way the numbers $6,7,8,9,10$ are distributed between the $a_i$ and $b_i$. If three are $b$s then two must be $a$s, for example.

  • $\begingroup$ Thank you Mark! I understand your solution very clearly. $\endgroup$ – Ice Inkberry Oct 6 '18 at 12:36

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