Imo 2010: Difference between revisions
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This is | This is the wiki page for the mini-polymath2 project, which seeks solutions to Question 5 of the 2010 International Mathematical Olympiad. | ||
The project will start at [http://www.timeanddate.com/worldclock/fixedtime.html?year=2010&month=7&day=8&hour=16&min=0&sec=0&p1=0 16:00 UTC July 8], and will be hosted at the [http://polymathprojects.org/ polymath blog]. A discussion thread will be hosted at [http://terrytao.wordpress.com Terry Tao's blog]. | The project will start at [http://www.timeanddate.com/worldclock/fixedtime.html?year=2010&month=7&day=8&hour=16&min=0&sec=0&p1=0 16:00 UTC July 8], and will be hosted at the [http://polymathprojects.org/ polymath blog]. A discussion thread will be hosted at [http://terrytao.wordpress.com Terry Tao's blog]. | ||
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== The question == | == The question == | ||
The question to be solved | The question to be solved is Question 5 of the [http://www.imo-official.org/problems.aspx 2010 International Mathematical Olympiad]: | ||
: '''Problem''' In each of six boxes <math>B_1, B_2, B_3, B_4, B_5, B_6</math> there is initially one coin. There are two types of operation allowed: | |||
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: ''Type 1:'' Choose a nonempty box <math>B_j</math> with <math>1 \leq j \leq 5</math>. Remove one coin from <math>B_j</math> and add two coins to <math>B_{j+1}</math>. | |||
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: ''Type 2:'' Choose a nonempty box <math>B_k</math> with <math>1 \leq k �\leq 4</math>. Remove one coin from <math>B_k</math> and exchange the contents of (possibly empty) boxes <math>B_{k+1}</math> and <math>B_{k+2}</math>. | |||
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: Determine whether there is a finite sequence of such operations that results in boxes <math>B_1, B_2, B_3, B_4, B_5</math> being empty and box <math>B_6</math> containing exactly <math>2010^{2010^2010}</math> coins. (Note that <math>a^{b^c} := a^{(b^c)}</math>.) | |||
== Observations and partial results == | == Observations and partial results == |
Revision as of 06:32, 8 July 2010
This is the wiki page for the mini-polymath2 project, which seeks solutions to Question 5 of the 2010 International Mathematical Olympiad.
The project will start at 16:00 UTC July 8, and will be hosted at the polymath blog. A discussion thread will be hosted at Terry Tao's blog.
Rules
This project will follow the usual polymath rules. In particular:
- Everyone is welcome to participate, though people who have already seen an external solution to the problem should probably refrain from giving spoilers throughout the experiment.
- This is a team effort, not a race between individuals. Rather than work for extended periods of time in isolation from the rest of the project, the idea is to come up with short observations (or to carry an observation of another participant further) and then report back what one gets to the rest of the team. Partial results or even failures can be worth reporting.
- Participants are encouraged to update the wiki, or to summarise progress within threads, for the benefit of others.
Threads
Discussion and planning:
- Future mini-polymath project: 2010 IMO Q6? June 12, 2010.
- Organising mini-polymath2 June 21, 2010.
- Mini-polymath2 start time, June 27, 2010.
Research:
- No research thread yet.
The question
The question to be solved is Question 5 of the 2010 International Mathematical Olympiad:
- Problem In each of six boxes [math]\displaystyle{ B_1, B_2, B_3, B_4, B_5, B_6 }[/math] there is initially one coin. There are two types of operation allowed:
- Type 1: Choose a nonempty box [math]\displaystyle{ B_j }[/math] with [math]\displaystyle{ 1 \leq j \leq 5 }[/math]. Remove one coin from [math]\displaystyle{ B_j }[/math] and add two coins to [math]\displaystyle{ B_{j+1} }[/math].
- Type 2: Choose a nonempty box [math]\displaystyle{ B_k }[/math] with [math]\displaystyle{ 1 \leq k �\leq 4 }[/math]. Remove one coin from [math]\displaystyle{ B_k }[/math] and exchange the contents of (possibly empty) boxes [math]\displaystyle{ B_{k+1} }[/math] and [math]\displaystyle{ B_{k+2} }[/math].
- Determine whether there is a finite sequence of such operations that results in boxes [math]\displaystyle{ B_1, B_2, B_3, B_4, B_5 }[/math] being empty and box [math]\displaystyle{ B_6 }[/math] containing exactly [math]\displaystyle{ 2010^{2010^2010} }[/math] coins. (Note that [math]\displaystyle{ a^{b^c} := a^{(b^c)} }[/math].)
Observations and partial results
- No partial results yet.
Possible strategies
- No possible strategies proposed yet.