Comments on: Minipolymath2 project: IMO 2010 Q5

By: Polymath project: social problem-solving | Civil Statistician

Polymath project: social problem-solving | Civil Statistician — Thu, 12 Jul 2012 18:31:07 +0000

[...] and the spirit behind it. Personally, I had a blast trying to contribute (if only a tiny bit) to the 2010 event. Dang, I almost had comment [...]

By: Mini-polymath 3: 2011 IMO question « What’s new

Mini-polymath 3: 2011 IMO question « What’s new — Fri, 10 Jun 2011 00:49:30 +0000

[...] The format of the last year’s mini-polymath project seemed to work well, so I am inclined to simply repeat that format without much modification this time around, in order to collect a consistent set of data about these projects. Thus, the project will start at a pre-arranged time and date, with plenty of advance notice, and be run simultaneously on three different sites: a “research thread” over at the polymath blog for the problem solving process, a “discussion thread” over at this blog for any meta-discussion about the project, and a wiki page at the polymath wiki to record the progress already made at the research thread. (Incidentally, there is a current discussion at the wiki about the logo for that site; please feel free to chip in your opinion on the various proposed icons.) The project will follow the usual polymath rules (as summarised for instance in the 2010 mini-polymath thread). [...]

By: Mini-polymath2 discussion thread « mathTHÍCHinTOÁNmyHỌCbrain

Mini-polymath2 discussion thread « mathTHÍCHinTOÁNmyHỌCbrain — Thu, 16 Dec 2010 12:43:10 +0000

[...] just opened the Mini-polymath2 project over at the polymath blog. I decided to use Q5 from the 2010 IMO in the end, rather than Q6, as [...]

By: IMO 2010 « Absolutely useless

IMO 2010 « Absolutely useless — Fri, 23 Jul 2010 20:34:39 +0000

[...] you can see in the above link, Tao started a mini-polymath project about problem 5. So here you can see how a group of mathematicians worked together to solve that problem (and see the problem [...]

By: Mark Bennet

Mark Bennet — Fri, 16 Jul 2010 21:13:07 +0000

I’ve been working on the biggest numbers you can get – based on what others have found.

I wonder – what is the most efficient solution – fewest number of moves?

By: Mark Bennet

Mark Bennet — Fri, 16 Jul 2010 17:16:55 +0000

There are various ways of getting a recursion formula going.

For n>1, let f(k,n,m) be the maximum value of d which can be obtained starting with

[n,m,0 ... 0] (with k-2 zeros)
and ending with
[0,d,0 ... 0]

Define f(1,n,0)=n (equivalent to [n] goes to [n])
Then (except perhaps for some marginal cases where m=1, which never arise staring with [n,0 ... 0]
for k>1 f(k,n,0)=f(k,n-1,2)
(for n>0) f(k,n,m) = f(k,n-1,f(k-1,m,0))
and f(k,0,m) = f(k-1,m,0)

By: Problema 5 de IMO 2010 resolvido colaborativamente no projecto Mini-Polymath2 « problemas | teoremas

Wed, 14 Jul 2010 22:22:43 +0000

[...] Problema Decorreu recentemente no The polymath blog e sob a organização de Terence Tao, a resolução conjunta do Problema 5 das Olimpíadas Internacionais de Matemática deste ano. Em vez do Problema 6, [...]

By: Kristal Cantwell

Kristal Cantwell — Tue, 13 Jul 2010 04:02:35 +0000

It looks like this can be generalized to give any finite number of coins in a finite number of boxes
has the total number of coins bounded.

By: Mark Bennet

Mark Bennet — Mon, 12 Jul 2010 22:02:51 +0000

Every extra place to the left adds an extra level of Knuth up-arrows to the maximum outcome of [n,0,0 ... ,0] provided that n is at least 2 – this is the function of the second type of move (see 47 above).

BUT with [1,?,?, ... ?] there can be advantages in using the first type of move more often, and the numbers in the left-hand places may not create the full value.

This creates constraints on how useful conserved quantities might be computed.

By: Mark Bennet

Mark Bennet — Mon, 12 Jul 2010 16:42:12 +0000

I think I can see the pattern here.

[1,1,1,1,1,1] goes to [1,0,0,2^14,0,0]
(nb the 7 is special at the tail – doesn’t quite fit the pattern)
(note 2^14 = 2^(2*7) which brings out the roles of 2 and 7 separately
and 2*2^(2^(2*7)) – two multiplications by 2, two powers
- the powers in between the multiplications in a chiastic structure)

[1,0,0,2^(2*7),0,0] goes to [0,2,0,2^(2*7),0,0]
The 2 in the second place is the source of two operations of the highest kind which appear in the answer.
The zeros at the end give space to repeat the power and multiplication operation in reverse.
This goes to [0,1,2^(2*7),0,0,0]
Which goes to [0,1,0,2^^(2^(2*7)),0,0]
Which goes to [0,0,2^^(2^(2*7)),0,0,0]
Which goes to [0,0,2^^(2^^(2^(2*7)),0,0]

… and eventually the result is 2*2^(2^^(2^^(2^(2*7)))