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[Math Puzzle] Day12 - Page 2

Blogs > evanthebouncy!
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King K. Rool
Profile Blog Joined May 2009
Canada4408 Posts
Last Edited: 2009-05-29 15:18:22
May 29 2009 15:17 GMT
#21
Psuedo_Utopia has the same solution I worked out, so if it IS that solution, then I don't think this is really undergrad stuff, more like high-school math contest TBH (if you did good on contests, not trying to come off as pretentious). Seems like problem solving skills are enough to arrive at the proof, and requires no theorems or anything.
mmp
Profile Blog Joined April 2009
United States2130 Posts
Last Edited: 2009-05-29 20:55:24
May 29 2009 20:14 GMT
#22
Yeah, it seems that everyone has the answer already. Here is my attempt at a semi-formal proof:

+ Show Spoiler +

Let a row contain an infinite sequence of elements belonging to {1,2,3,4,5,6}. A row is guaranteed to contain at least 2 of some element (since there are greater than 6 elements in the sequence). Let N be the width that separates these two identical elements in a row.

Define the column to be an infinite sequence of rows. No two rows in a column can be identical, nor can any two adjacent sub-sequences (both) containing identical elements be identical, lest there exist a rectangle.

An N-wide sub-sequence containing identical elements must exist and has only 6^N permutations. There are infinite unique rows in the column, but not infinite permutations of this sub-sequence.
I (λ (foo) (and (<3 foo) ( T_T foo) (RAGE foo) )) Starcraft
evanthebouncy!
Profile Blog Joined June 2006
United States12796 Posts
June 01 2009 07:48 GMT
#23
On May 30 2009 00:17 King K. Rool wrote:
Psuedo_Utopia has the same solution I worked out, so if it IS that solution, then I don't think this is really undergrad stuff, more like high-school math contest TBH (if you did good on contests, not trying to come off as pretentious). Seems like problem solving skills are enough to arrive at the proof, and requires no theorems or anything.


right and you come to a thread titled "math puzzle" expecting things that requires stacked theorems. This problem is in chapter 1 of my Combinatorics book on pigeon hole principle.
Life is run, it is dance, it is fast, passionate and BAM!, you dance and sing and booze while you can for now is the time and time is mine. Smile and laugh when still can for now is the time and soon you die!
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