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# Colouring watch

1. http://www.bmoc.maths.org/home/bmo1-2005.pdf

This concerns q4 on the paper.
My solution is as follows.

Let S(n) denote the statement in the question.
Then it is clear that S(i) true implies S(j) true for all j not less than i.

Claim 1: S(10) is false
Claim 2: S(11) is true

If we can prove these 2 claims, then n=11 is the least natural number such that S(n) is true (since if r<11 has S(r) true then S(10) is true by the comment above, and this contradicts claim 1).

Claim 1: The colouring R={1,2,9,10} B={3,4,5,6,7,8} shows this (after a simple check which I can't be bothered to write out).

Claim 2: Suppose otherwise. WLOG this colouring has 1=R, 3=B, 9=R,
11=B (where 1=R means 1 is in the set R).

Case a) 2=R.
Then we must have 4=B. But then 3,4,11 are in B and 3+4+4=11, a contradiction.

Case b) 2=B.
Then we have that 6=R and 8=R. But then 1,6,8 are in R and 1+1+6=8, a contradiction.

Done.

Does everyone agree with this result and proof or does it need more clarification to get full marks in the BMO / convince people it's correct.

Also, I thought this would have some sort of 'pigeonhole principle' type approach, rather than the mundane thing I've done above. Does anyone have another solution to this?

Cheers
2. bump

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