Thalia
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So 6/7 in the last post and check CEX?
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Think I found one of the 2 I changed from correct to incorrect. But I think that would leave one of these as still wrong though I can't figure out which. . .
SpoilerMMM (2) 5.11.23 5.11.15 CCC (3) 1.3.7 1.3.27 1.9.21 CCX (3) 1.3.14 1.9.14 1.14.27 CEX (4) 1.2.14 1.6.14 1.8.14 1.14.18 CMX (2) 1.5.14 5.14.21 EMX (3) 2.5.14 5.12.14 5.14.20 MMX (2) 5.11.14 5.14.23 -
Another stab at it.
SpoilerMMM (2) 5.11.23 5.11.15 CCC (2) 1.3.7 1.3.27 CCX (3) 1.3.14 1.9.14 1.14.27 CEX (4) 1.2.14 1.6.14 1.8.14 1.14.18 CMX (2) 1.5.14 5.14.21 EEX (6) 2.4.14 2.6.14 2.8.14 2.14.16 2.14.18 2.14.26 EMX (3) 2.5.14 5.12.14 5.14.20 MMX (2) 5.11.14 5.14.23 CCM (8) 1.3.5 1.5.9 1.5.19 1.5.21 1.5.25 1.5.27 5.19.21 5.19.27 CMM (5) 1.5.11 5.7.11 5.9.11 5.11.25 1.5.23 EMM (8) 2.5.11 4.5.11 5.6.11 5.8.11 5.11.18 5.11.24 2.5.23 5.10.23 EEE (11) 2.4.6 2.4.10 2.4.12 2.4.16 2.4.18 2.4.20 2.4.22 2.4.24 2.4.26 2.8.20 2.8.22 EEM (18) 2.4.5 2.5.8 2.5.10 2.5.12 2.5.16 2.5.18 2.5.20 2.5.22 2.5.24 2.5.26 5.10.12 5.10.16 5.10.20 5.10.22 5.10.24 5.10.26 5.20.22 5.20.26 CCE (15) 1.2.3 1.2.7 1.2.9 1.2.19 1.2.21 1.2.25 2.7.9 2.7.19 2.7.21 2.7.25 2.7.27 2.9.19 2.9.21 2.9.25 2.9.27 CEE (21) 1.2.4 1.2.6 1.2.8 1.2.12 1.2.16 1.2.18 1.2.20 1.2.22 1.2.24 1.2.26 1.6.8 1.6.12 1.6.16 1.6.18 1.6.22 1.6.24 1.6.26 1.8.12 1.8.18 1.8.24 1.8.26 CEM (24) 1.2.5 1.4.5 1.5.6 1.5.8 1.5.10 1.5.12 1.5.16 1.5.18 1.5.20 1.5.22 1.5.24 1.5.26 2.5.19 4.5.19 5.6.19 5.8.19 5.10.19 5.12.19 5.16.19 5.18.19 5.19.20 5.19.22 5.19.24 5.19.26 New total is 134. At least one of those is wrong though according to having 6/16 correct the first time...
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Revision to CCC and EEX with numbers.
SpoilerMMM (2) 5.11.23 5.11.15 CCC (2) 1.3.7 1.3.27 CCX (3) 1.3.14 1.9.14 1.14.27 CEX (4) 1.2.14 1.6.14 1.8.14 1.14.18 CMX (2) 1.5.14 5.14.21 EEX (6) 2.4.14 2.6.14 2.8.14 2.14.16 2.14.18 2.14.26 EMX (3) 2.5.14 5.12.14 5.14.20 MMX (2) 5.11.14 5.14.23 -
4 hours ago, Molly Mae said:
For MMM, you can have the middle out of opposite sides and a middle between them or you can have a middle out of the top of the cube and two out of sides that are next to each other. I will provide the numbers using CaptainEd's system later.
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Pretty sure these counts are wildly inaccurate but here goes...:
SpoilerC C C 3 C C E 21 C C M 5 C E E 24 C E M 24 C M M 6 E E E 12 E E M 20 E M M 10 M M M 2 C C X 3 C E X 4 C M X 2 E E X 5 E M X 3 M M X 2 146 -
For Bert, can you draw lines between points of intersection (like the theorem rocdocmac mentioned) or does the triangle have to be formed from the trisecting lines as attached?
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Welcome back Cavenglok!
That series sounds vaguely familiar but I don't think I followed it at the time. It sounds interesting though.
Good to see another former member back.
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1 hour ago, rocdocmac said:
Not the most elegant sketch, but here's one way to plant 12 trees in 18 rows of 3 each.
Thus, if Al and Chuck told the truth and Bert tried (but failed) to make a sketch for the inspector (none of the trisectors colinear), then Dick is the third truth teller. So far all four (living) brothers already got the blame, but it now appears that Bert is in trouble! Bert also mentioned that he "was analyzing random polygons with 3 sides" ... why didn't he say triangles?
I thought that was weird at first too but I attributed it to the Threedie Bros obsession with 3's. I found a picture of a trisected equilateral triangle. None of the triangles with a corner at the trisected angles work because they're only 20 or 40 degrees at the corner. There's 6 smaller more or less identical triangles around the center. I haven't worked out the math but one of the angles is nearly 90 degrees so those are out too. So starting with an equilateral triangle seems to disprove Bert's statement unless he has unusual luck in picking random triangles or you count the original triangle.
I am curious about bonanova's 8 statements comment though. 2 statements per person?
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Can't prove it yet but the triangle still seems off. Equilateral was an attempt at checking. Not meant to cover all possibilities.
I'm not sure what rocdocmac proved other than adding up a third of the angles of a triangle gives you 60 degrees?
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If I understand right, Bert's statement doesn't hold up starting with an equilateral triangle. But does making a false statement necessarily mean guilt or could it be a mistake on their part?
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There are 12 edges. According to CaptainEd's numbering system, they would be the even cubelets with the exception of 14 (the core). I used cubelet 1 as the corner. 2 and 4 make the same shape. As far as I can tell, 6 and 8 are 2 more. Cubelet 1 is part of 3 faces of the cube so any of the edges on those faces would make one of those 3 shapes. 10, 12, 16, 20, and 22 are out for that reason. 18 gives a 4th shape. 24 and 26 are the same as 18.
So 4 shapes for edge/corner. New total is 22? I'm wishing I had a real cube I could take apart now...
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After recounting...
I think CaptainEd had it right but forgot to count the possibilities with the core.
24
3 possibilities if the core is removed
2E 5
2M 2
2C 3
EM 3
EC 6
MC 2 -
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Make that 25. 4 possibilities for edges.
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23?
I get 3 possibilities if the core is removed.
Corner (C)
Edge (E)
Middle (M)
2E 2
2M 2
2C 3
EM 4
EC 7
MC 2 -
Without assuming a minimum number of peas in her hand, slightly greater than 1/2. There are two possibilities. You have an even number of peas or an odd number of peas (in the bag).
If you have an even number, the probability that she pulled out an odd number is 1/2.
If you have an odd number, there is one more odd number than even. So the probability is greater than 1/2.
Average the two and it's greater than 1/2.
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Without an outside person
Get some containers you can't see through with a slot. Sort of like a mini ballot box. Mark one for each number place (ones, tens, hundreds, etc). Use something cheap like paper clips and place them in the containers according to your salary. So if your salary is $36,000, you would place 3 paperclips in the 10,000 box and 6 in the 1,000 box. Since no one knows who put which paperclips in each box, you can add up the paperclips and average them.
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Sounds like a seagull. Don't know about actors though. That line makes me think goose/ gosling. Don't know about the beach though.
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I wouldn't mind one.
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Toy Story? Maybe the 3rd?
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1
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Not sure I understand what you're getting at so this probably isn't what you're going for but I'd be recommending protecting those sections more...
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If 385 is the LCM, I would assume you could just double it and add 1. So after 1 and 386, 771?
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...I interpreted "have another go" as meaning more was possible... Guess I should have tried harder to spoiler.
For spoilers on phone, use (spoiler)your text here(/spoiler) but use [ ] instead of ( ).
...I interpreted "have another go" as meaning more was possible... Guess I should have tried harder to spoiler.
For spoilers on phone, use (spoiler)your text here(/spoiler) but use [ ] instead of ( ).
Cubicle Stack #2
in New Logic/Math Puzzles
Posted
I see what you mean about CEX although I'm not sure how to show that with the numbering system. That throws a wrench in my counting. . .