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#1 |
"Matthew Anderson"
Dec 2010
Oregon, USA
2×3×7×17 Posts |
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I'm tottally into eazy puzzles. Check out this one. It involves a twisting motion.
Regards Matthew |
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#2 |
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
41·229 Posts |
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Hanayama Quartet is pretty good.
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#3 |
"Matthew Anderson"
Dec 2010
Oregon, USA
2×3×7×17 Posts |
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#4 |
"Mark"
Apr 2003
Between here and the
61×103 Posts |
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This is about half of my collection. At one time I considered taking a bunch to work and requiring anyone who had a question of me to solve a puzzle first because I am one of the "go to" people whenever someone has a question. I quickly realized that I would either scare people away or that people would end up wasting their time trying to solve these puzzles as opposed to doing what they are paid to do.
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#5 |
Jul 2003
wear a mask
62316 Posts |
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I found a solution to Aristotle's number puzzle:
Not exactly spoiler-free description here My solution came about by sheer dumb luck. I noticed there are 4 types of pieces: a central interior piece, 6 non-central interior pieces, 6 exterior corners and 6 non-corner exterior pieces. While trying to place a bound on the possible values of one of the types, I tried an iteration that kept allowing me to place more pieces until the problem was solved! This puzzle is a real treat! Several times I was tempted to write a program to brute-force (or maybe a little more elegantly) search for a solution, but I'm glad I persisted in solving it by hand. Last fiddled with by masser on 2021-04-03 at 06:53 |
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#6 |
Romulan Interpreter
Jun 2011
Thailand
33·347 Posts |
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I remember that was quite easy. Write in a table in Excel (by hand) all possibilities to make 38 with 3 numbers, like 1, 18, 19; 2, 17, 19; 3, 16, 19; 3, 17, 18; 4, 15, 19; 4, 16, 18; 5, 14, 19; 5, 15, 18; 5, 16, 17; etc, which is very easy, they are only 28, and there is obviously only one way to chain them to have the common corners around being part of two sums (they are 30, but you have to consider that 1 and 2 can't be part of any triple, because they appear only once, so they can't be corners, and in the same time then must appear on the sums of 5 - why? - so this tells you that 1 and 2 are inside, 19 and 18 are outside, 19 can't be a corner, therefore 18 must be one). At the time I did that, I think it took me less than it took to some other guy to write a search program that would need few seconds or a minute to solve it. I remember also that my struggle was related to the fact that I (wrongly) reached the conclusion that 10 has to be in the middle, I don't know why, probably based on magic squares
![]() Last fiddled with by LaurV on 2021-04-03 at 06:50 |
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#7 | |
Sep 2017
1438 Posts |
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