An IMO Problem With A Bizarre Unintuitive Solution

Hi,

In this video I'll be solving problem 5 from the 1983 IMO (International Mathematical Olympiad). We want to know if there exists a set of 1983 positive integers ≤100,000 such that no 3 are consecutive terms in an arithmetic progression.

I do recommend you have a go at the problem (I got stuck, but I'm sure you can power through). Let me know how you got on in the comments!

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And, as always, any comments, feedback or suggestions are welcomed!

Thanks.

[thirunarayanansampat]

Your daily videos have a becme a source of my entertainments. Feel like you are a math encyclopedia. Awesome and Keep it up
.. Can you please suggest a good math book which can be a pure entertainment to solve. ?

[spazmoidectomorf]

I had tried this problem and would like to explain some of the thought process.
Let x+z=2y
Meaning it has to be an arithmetic sequence
It may seem confusing to switch from base 2 to 3. If you think about it you can represent any number from 0 to 2047 in an 11 digit binary code. Now think about base 3: just writing a couple of examples, 1, 3, 4, 9, 10, 12, 13, 27, 28, 30, 31, 36 you sort of see the way the sequence jumps to never produce an arithmetic progression (which I will not explain here). Now, the largest number expressable in base 3 is which is 88753.

So now the final part, you say that both x and z belong to the set of the base 3 numbers (essentially saying that they have no sort of progression) but y must be expressed as in terms of 2's ands 0's. In base 2 you could simply put a 0 and put a 1 in the column before, but this is base 3 so it has to remain a 2.

[particleonazock]

Daily Upload Streak: 2. Not sure if this has been broken before, please correct if wrong