tojo928
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Posts posted by tojo928
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Think I have the ideal solution here
SpoilerA (0,0); B (1,0); C(1,1); D (0,1); E (cos(30),sin(30)); F (cos(60),sin(60))
This solution assumes if you are excatly the same distance from points you get to choose. Very slight tweaks to the points can force the longest path
Start at A. B,D,E&F are all 1 mile away so pick E (+1 mile)
At E. B and F are both same distance away (2 sin(15)~.5176) so pick B (+.5176 mile)
At B. A,E,&F are all 1 mile away so pick F (+1 mile)
At F, D is closest (2 sin(15)) (+.5176 mile)
At D, C is closest (+1 mile)
At C, return to A (2^.5~1.414
Total distance walked= 3*(1)+2*(2sin(15))+2^.5~5.449
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The first number is the lowest possible answer, the dots represent a series of numbers that fit the sequence. Give the 3 blanks in the sequence. You might have to think alittle off the wall *wink* to come up with the pattern.
2, 3, 5, 20, 26.......... 929, 930, 936, 939, 948, 950, 956, ______, _______, ______....2038, 2058.....
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Spoiler
The number is 840
Start with number 2. Must be True. That makes number 1 False. Which makes 7 and 8 False. With 7 False we know the number is <=1000. With 8 False we know there are exactly 5 True and 5 False statements
So Far we have 1)F 2)T 3) 4) 5) 6) 7)F 8)F 9) 10)
Now assume number 3 is true. If the number is prime, then 4, 5, 6, and 9 must be false. That would mean that there are 7 false statements which contradicts the conclusion about number 8. Therefore 3 must be False.
Now we have 1)F 2)T 3)F 4) 5) 6) 7)F 8)F 9) 10)
Next we look at statement 10. Assume it is True. This makes 9 False and 4,5,6 True. The number would be defined as <1000 (statement 7), has 18 as divisor (statement 4), has >27 divisors(statement 5 {2+4+5+6+10}), has 4 prime divisors (statement 6), and divisor is not a cube (statement 9). The only number divisible by 18 that has more than 27 factors is 720. This is divisible by 8 and therefore can not be true. So statement 10 must be False.
This results in the statements as follows 1)F 2)T 3)F 4)T 5)T 6)T 7)F 8)F 9)T 10)F
So the number must be <1000, divisible by 20, divisible by a cube, have 4 prime divisors, and have >26 divisors. The only number that fits this is 840.
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Still working on the second question
Second robot not broken
SpoilerRobots 2,1,1,1,1
Robot 1 2 3 4 5 6 7 8 2 4 3 2 1 8 7 6 5 1 3 2 1 8 7 6 5 4 1 2 1 8 7 6 5 4 3 1 1 8 7 6 5 4 3 2 1 8 7 6 5 4 3 2 1 -
my bad. I left a couple options for the exterior primes off my list.
On 11/3/2016 at 9:52 PM, jasen said:It is possible, I have it. I will show it next week if nobody find it until 10 Nov 2016.
Spoiler3 1 1
1 5 7
1 7 9
Fun extra question to add in to challenge on this one. What is the lowest number of unique primes that could be used in this and how many unique ways (including rotation and reflection) can this be made?
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Spoiler
Not possible to get 9 unique values
Most you can get is 8 with this(and all the unique rotations and reflections):
7 3 9
3 8/5 3
3 3 7
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Spoiler
1,19,3,12,16,14,18,10,7,6,15,2,5,9,13,17,8,11,4
1,19,3,18,16,14,12,4,13,6,9,8,17,15,7,5,2,11,10
Set a, b, and c to 1, 19, and 3 to constrain the only way 19 can fit with the requirements. Used excel to solve for the remaining corners and center.
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Spoiler
Take number 27 (the middle road)
The first 4 villagers can not be "useless" based on the possible outcomes of their clues. This leaves the remaining villagers. Only 1 combination of numbers leaves 3 of them giving useless information (15, 27, and 39 would have villagers 7, 8, and 9 giving useless information. From there its just a matter of testing the remaining outcomes.
15 27 39 1 W R W 2 R W W 3 W R W 4 W W R 5 R W R 6 R W R 7 U U U 8 U U U 9 U U U 10 W R W 11 W R W -
Spoiler
I got 32sqrt(3)-48+54/sqrt(3)
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Spoiler
Firstly, there's way to many rectangles so I'm going to logic some away. Any time a rectangle shares the same full line segment that segment can either be an integer or not. By that logic, the other side of the rectangle would be the opposite. This means the combination of these 2 rectangles has the same properties as the 2 smaller rectangles. You can use this logic multiple times and end up with only 5 rectangles that look roughly like this
Now the figure is effectively rotationally congruent so we only need to prove one side full side in an integer and the proof will hold for the other possibilities. So both line segments that make up the right side can be integers, on is and one isn't, and both aren't.
Both integers obviously proves the theory for that case
One integer and one not creates a situation where the center rectangle will either have to both sides be integers or both side not be integers which contradicts the problem requirements.
Both non-integers creates a situation where both of the top have to be integers and proves the theory.
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I can move the good Dr. for less than $25
SpoilerIn 324 total steps the Dr. can visit every vertex for no more than $12.60 going either way.
Method: From the starting point move in one direction (say Clockwise), then move across, then move in the other direction (CCW) until moving across would bring you to a new location. Then move across and begin moving in the original direction until you reach a new vertex. Repeat until completed.
Data. Assume starting at #1 and all vertices are number around clockwise from here.
SpoilerStart Direction End Cost Running total 1 CW 2 $ 1.05 $ 1.05 2 A 27 $ - $ 1.05 27 CCW 26 $ (1.05) $ - 26 CCW 25 $ (1.05) $ (1.05) 25 A 50 $ - $ (1.05) 50 CW 1 $ 1.05 $ - 1 CW 2 $ 1.05 $ 1.05 2 CW 3 $ 1.05 $ 2.10 3 A 28 $ - $ 2.10 28 CCW 27 $ (1.05) $ 1.05 27 CCW 26 $ (1.05) $ - 26 CCW 25 $ (1.05) $ (1.05) 25 CCW 24 $ (1.05) $ (2.10) 24 A 49 $ - $ (2.10) 49 CW 50 $ 1.05 $ (1.05) 50 CW 1 $ 1.05 $ - 1 CW 2 $ 1.05 $ 1.05 2 CW 3 $ 1.05 $ 2.10 3 CW 4 $ 1.05 $ 3.15 4 A 29 $ - $ 3.15 29 CCW 28 $ (1.05) $ 2.10 28 CCW 27 $ (1.05) $ 1.05 27 CCW 26 $ (1.05) $ - 26 CCW 25 $ (1.05) $ (1.05) 25 CCW 24 $ (1.05) $ (2.10) 24 CCW 23 $ (1.05) $ (3.15) 23 A 48 $ - $ (3.15) 48 CW 49 $ 1.05 $ (2.10) 49 CW 50 $ 1.05 $ (1.05) 50 CW 1 $ 1.05 $ - 1 CW 2 $ 1.05 $ 1.05 2 CW 3 $ 1.05 $ 2.10 3 CW 4 $ 1.05 $ 3.15 4 CW 5 $ 1.05 $ 4.20 5 A 30 $ - $ 4.20 30 CCW 29 $ (1.05) $ 3.15 29 CCW 28 $ (1.05) $ 2.10 28 CCW 27 $ (1.05) $ 1.05 27 CCW 26 $ (1.05) $ - 26 CCW 25 $ (1.05) $ (1.05) 25 CCW 24 $ (1.05) $ (2.10) 24 CCW 23 $ (1.05) $ (3.15) 23 CCW 22 $ (1.05) $ (4.20) 22 A 47 $ - $ (4.20) 47 CW 48 $ 1.05 $ (3.15) 48 CW 49 $ 1.05 $ (2.10) 49 CW 50 $ 1.05 $ (1.05) 50 CW 1 $ 1.05 $ - 1 CW 2 $ 1.05 $ 1.05 2 CW 3 $ 1.05 $ 2.10 3 CW 4 $ 1.05 $ 3.15 4 CW 5 $ 1.05 $ 4.20 5 CW 6 $ 1.05 $ 5.25 6 A 31 $ - $ 5.25 31 CCW 30 $ (1.05) $ 4.20 30 CCW 29 $ (1.05) $ 3.15 29 CCW 28 $ (1.05) $ 2.10 28 CCW 27 $ (1.05) $ 1.05 27 CCW 26 $ (1.05) $ - 26 CCW 25 $ (1.05) $ (1.05) 25 CCW 24 $ (1.05) $ (2.10) 24 CCW 23 $ (1.05) $ (3.15) 23 CCW 22 $ (1.05) $ (4.20) 22 CCW 21 $ (1.05) $ (5.25) 21 A 46 $ - $ (5.25) 46 CW 47 $ 1.05 $ (4.20) 47 CW 48 $ 1.05 $ (3.15) 48 CW 49 $ 1.05 $ (2.10) 49 CW 50 $ 1.05 $ (1.05) 50 CW 1 $ 1.05 $ - 1 CW 2 $ 1.05 $ 1.05 2 CW 3 $ 1.05 $ 2.10 3 CW 4 $ 1.05 $ 3.15 4 CW 5 $ 1.05 $ 4.20 5 CW 6 $ 1.05 $ 5.25 6 CW 7 $ 1.05 $ 6.30 7 A 32 $ - $ 6.30 32 CCW 31 $ (1.05) $ 5.25 31 CCW 30 $ (1.05) $ 4.20 30 CCW 29 $ (1.05) $ 3.15 29 CCW 28 $ (1.05) $ 2.10 28 CCW 27 $ (1.05) $ 1.05 27 CCW 26 $ (1.05) $ - 26 CCW 25 $ (1.05) $ (1.05) 25 CCW 24 $ (1.05) $ (2.10) 24 CCW 23 $ (1.05) $ (3.15) 23 CCW 22 $ (1.05) $ (4.20) 22 CCW 21 $ (1.05) $ (5.25) 21 CCW 20 $ (1.05) $ (6.30) 20 A 45 $ - $ (6.30) 45 CW 46 $ 1.05 $ (5.25) 46 CW 47 $ 1.05 $ (4.20) 47 CW 48 $ 1.05 $ (3.15) 48 CW 49 $ 1.05 $ (2.10) 49 CW 50 $ 1.05 $ (1.05) 50 CW 1 $ 1.05 $ - 1 CW 2 $ 1.05 $ 1.05 2 CW 3 $ 1.05 $ 2.10 3 CW 4 $ 1.05 $ 3.15 4 CW 5 $ 1.05 $ 4.20 5 CW 6 $ 1.05 $ 5.25 6 CW 7 $ 1.05 $ 6.30 7 CW 8 $ 1.05 $ 7.35 8 A 33 $ - $ 7.35 33 CCW 32 $ (1.05) $ 6.30 32 CCW 31 $ (1.05) $ 5.25 31 CCW 30 $ (1.05) $ 4.20 30 CCW 29 $ (1.05) $ 3.15 29 CCW 28 $ (1.05) $ 2.10 28 CCW 27 $ (1.05) $ 1.05 27 CCW 26 $ (1.05) $ - 26 CCW 25 $ (1.05) $ (1.05) 25 CCW 24 $ (1.05) $ (2.10) 24 CCW 23 $ (1.05) $ (3.15) 23 CCW 22 $ (1.05) $ (4.20) 22 CCW 21 $ (1.05) $ (5.25) 21 CCW 20 $ (1.05) $ (6.30) 20 CCW 19 $ (1.05) $ (7.35) 19 A 44 $ - $ (7.35) 44 CW 45 $ 1.05 $ (6.30) 45 CW 46 $ 1.05 $ (5.25) 46 CW 47 $ 1.05 $ (4.20) 47 CW 48 $ 1.05 $ (3.15) 48 CW 49 $ 1.05 $ (2.10) 49 CW 50 $ 1.05 $ (1.05) 50 CW 1 $ 1.05 $ - 1 CW 2 $ 1.05 $ 1.05 2 CW 3 $ 1.05 $ 2.10 3 CW 4 $ 1.05 $ 3.15 4 CW 5 $ 1.05 $ 4.20 5 CW 6 $ 1.05 $ 5.25 6 CW 7 $ 1.05 $ 6.30 7 CW 8 $ 1.05 $ 7.35 8 CW 9 $ 1.05 $ 8.40 9 A 34 $ - $ 8.40 34 CCW 33 $ (1.05) $ 7.35 33 CCW 32 $ (1.05) $ 6.30 32 CCW 31 $ (1.05) $ 5.25 31 CCW 30 $ (1.05) $ 4.20 30 CCW 29 $ (1.05) $ 3.15 29 CCW 28 $ (1.05) $ 2.10 28 CCW 27 $ (1.05) $ 1.05 27 CCW 26 $ (1.05) $ - 26 CCW 25 $ (1.05) $ (1.05) 25 CCW 24 $ (1.05) $ (2.10) 24 CCW 23 $ (1.05) $ (3.15) 23 CCW 22 $ (1.05) $ (4.20) 22 CCW 21 $ (1.05) $ (5.25) 21 CCW 20 $ (1.05) $ (6.30) 20 CCW 19 $ (1.05) $ (7.35) 19 CCW 18 $ (1.05) $ (8.40) 18 A 43 $ - $ (8.40) 43 CW 44 $ 1.05 $ (7.35) 44 CW 45 $ 1.05 $ (6.30) 45 CW 46 $ 1.05 $ (5.25) 46 CW 47 $ 1.05 $ (4.20) 47 CW 48 $ 1.05 $ (3.15) 48 CW 49 $ 1.05 $ (2.10) 49 CW 50 $ 1.05 $ (1.05) 50 CW 1 $ 1.05 $ - 1 CW 2 $ 1.05 $ 1.05 2 CW 3 $ 1.05 $ 2.10 3 CW 4 $ 1.05 $ 3.15 4 CW 5 $ 1.05 $ 4.20 5 CW 6 $ 1.05 $ 5.25 6 CW 7 $ 1.05 $ 6.30 7 CW 8 $ 1.05 $ 7.35 8 CW 9 $ 1.05 $ 8.40 9 CW 10 $ 1.05 $ 9.45 10 A 35 $ - $ 9.45 35 CCW 34 $ (1.05) $ 8.40 34 CCW 33 $ (1.05) $ 7.35 33 CCW 32 $ (1.05) $ 6.30 32 CCW 31 $ (1.05) $ 5.25 31 CCW 30 $ (1.05) $ 4.20 30 CCW 29 $ (1.05) $ 3.15 29 CCW 28 $ (1.05) $ 2.10 28 CCW 27 $ (1.05) $ 1.05 27 CCW 26 $ (1.05) $ - 26 CCW 25 $ (1.05) $ (1.05) 25 CCW 24 $ (1.05) $ (2.10) 24 CCW 23 $ (1.05) $ (3.15) 23 CCW 22 $ (1.05) $ (4.20) 22 CCW 21 $ (1.05) $ (5.25) 21 CCW 20 $ (1.05) $ (6.30) 20 CCW 19 $ (1.05) $ (7.35) 19 CCW 18 $ (1.05) $ (8.40) 18 CCW 17 $ (1.05) $ (9.45) 17 A 42 $ - $ (9.45) 42 CW 43 $ 1.05 $ (8.40) 43 CW 44 $ 1.05 $ (7.35) 44 CW 45 $ 1.05 $ (6.30) 45 CW 46 $ 1.05 $ (5.25) 46 CW 47 $ 1.05 $ (4.20) 47 CW 48 $ 1.05 $ (3.15) 48 CW 49 $ 1.05 $ (2.10) 49 CW 50 $ 1.05 $ (1.05) 50 CW 1 $ 1.05 $ - 1 CW 2 $ 1.05 $ 1.05 2 CW 3 $ 1.05 $ 2.10 3 CW 4 $ 1.05 $ 3.15 4 CW 5 $ 1.05 $ 4.20 5 CW 6 $ 1.05 $ 5.25 6 CW 7 $ 1.05 $ 6.30 7 CW 8 $ 1.05 $ 7.35 8 CW 9 $ 1.05 $ 8.40 9 CW 10 $ 1.05 $ 9.45 10 CW 11 $ 1.05 $ 10.50 11 A 36 $ - $ 10.50 36 CCW 35 $ (1.05) $ 9.45 35 CCW 34 $ (1.05) $ 8.40 34 CCW 33 $ (1.05) $ 7.35 33 CCW 32 $ (1.05) $ 6.30 32 CCW 31 $ (1.05) $ 5.25 31 CCW 30 $ (1.05) $ 4.20 30 CCW 29 $ (1.05) $ 3.15 29 CCW 28 $ (1.05) $ 2.10 28 CCW 27 $ (1.05) $ 1.05 27 CCW 26 $ (1.05) $ - 26 CCW 25 $ (1.05) $ (1.05) 25 CCW 24 $ (1.05) $ (2.10) 24 CCW 23 $ (1.05) $ (3.15) 23 CCW 22 $ (1.05) $ (4.20) 22 CCW 21 $ (1.05) $ (5.25) 21 CCW 20 $ (1.05) $ (6.30) 20 CCW 19 $ (1.05) $ (7.35) 19 CCW 18 $ (1.05) $ (8.40) 18 CCW 17 $ (1.05) $ (9.45) 17 CCW 16 $ (1.05) $ (10.50) 16 A 41 $ - $ (10.50) 41 CW 42 $ 1.05 $ (9.45) 42 CW 43 $ 1.05 $ (8.40) 43 CW 44 $ 1.05 $ (7.35) 44 CW 45 $ 1.05 $ (6.30) 45 CW 46 $ 1.05 $ (5.25) 46 CW 47 $ 1.05 $ (4.20) 47 CW 48 $ 1.05 $ (3.15) 48 CW 49 $ 1.05 $ (2.10) 49 CW 50 $ 1.05 $ (1.05) 50 CW 1 $ 1.05 $ - 1 CW 2 $ 1.05 $ 1.05 2 CW 3 $ 1.05 $ 2.10 3 CW 4 $ 1.05 $ 3.15 4 CW 5 $ 1.05 $ 4.20 5 CW 6 $ 1.05 $ 5.25 6 CW 7 $ 1.05 $ 6.30 7 CW 8 $ 1.05 $ 7.35 8 CW 9 $ 1.05 $ 8.40 9 CW 10 $ 1.05 $ 9.45 10 CW 11 $ 1.05 $ 10.50 11 CW 12 $ 1.05 $ 11.55 12 A 37 $ - $ 11.55 37 CCW 36 $ (1.05) $ 10.50 36 CCW 35 $ (1.05) $ 9.45 35 CCW 34 $ (1.05) $ 8.40 34 CCW 33 $ (1.05) $ 7.35 33 CCW 32 $ (1.05) $ 6.30 32 CCW 31 $ (1.05) $ 5.25 31 CCW 30 $ (1.05) $ 4.20 30 CCW 29 $ (1.05) $ 3.15 29 CCW 28 $ (1.05) $ 2.10 28 CCW 27 $ (1.05) $ 1.05 27 CCW 26 $ (1.05) $ - 26 CCW 25 $ (1.05) $ (1.05) 25 CCW 24 $ (1.05) $ (2.10) 24 CCW 23 $ (1.05) $ (3.15) 23 CCW 22 $ (1.05) $ (4.20) 22 CCW 21 $ (1.05) $ (5.25) 21 CCW 20 $ (1.05) $ (6.30) 20 CCW 19 $ (1.05) $ (7.35) 19 CCW 18 $ (1.05) $ (8.40) 18 CCW 17 $ (1.05) $ (9.45) 17 CCW 16 $ (1.05) $ (10.50) 16 CCW 15 $ (1.05) $ (11.55) 15 A 40 $ - $ (11.55) 40 CW 41 $ 1.05 $ (10.50) 41 CW 42 $ 1.05 $ (9.45) 42 CW 43 $ 1.05 $ (8.40) 43 CW 44 $ 1.05 $ (7.35) 44 CW 45 $ 1.05 $ (6.30) 45 CW 46 $ 1.05 $ (5.25) 46 CW 47 $ 1.05 $ (4.20) 47 CW 48 $ 1.05 $ (3.15) 48 CW 49 $ 1.05 $ (2.10) 49 CW 50 $ 1.05 $ (1.05) 50 CW 1 $ 1.05 $ - 1 CW 2 $ 1.05 $ 1.05 2 CW 3 $ 1.05 $ 2.10 3 CW 4 $ 1.05 $ 3.15 4 CW 5 $ 1.05 $ 4.20 5 CW 6 $ 1.05 $ 5.25 6 CW 7 $ 1.05 $ 6.30 7 CW 8 $ 1.05 $ 7.35 8 CW 9 $ 1.05 $ 8.40 9 CW 10 $ 1.05 $ 9.45 10 CW 11 $ 1.05 $ 10.50 11 CW 12 $ 1.05 $ 11.55 12 CW 13 $ 1.05 $ 12.60 13 A 38 $ - $ 12.60 38 CCW 37 $ (1.05) $ 11.55 37 CCW 36 $ (1.05) $ 10.50 36 CCW 35 $ (1.05) $ 9.45 35 CCW 34 $ (1.05) $ 8.40 34 CCW 33 $ (1.05) $ 7.35 33 CCW 32 $ (1.05) $ 6.30 32 CCW 31 $ (1.05) $ 5.25 31 CCW 30 $ (1.05) $ 4.20 30 CCW 29 $ (1.05) $ 3.15 29 CCW 28 $ (1.05) $ 2.10 28 CCW 27 $ (1.05) $ 1.05 27 CCW 26 $ (1.05) $ - 26 CCW 25 $ (1.05) $ (1.05) 25 CCW 24 $ (1.05) $ (2.10) 24 CCW 23 $ (1.05) $ (3.15) 23 CCW 22 $ (1.05) $ (4.20) 22 CCW 21 $ (1.05) $ (5.25) 21 CCW 20 $ (1.05) $ (6.30) 20 CCW 19 $ (1.05) $ (7.35) 19 CCW 18 $ (1.05) $ (8.40) 18 CCW 17 $ (1.05) $ (9.45) 17 CCW 16 $ (1.05) $ (10.50) 16 CCW 15 $ (1.05) $ (11.55) 15 CCW 14 $ (1.05) $ (12.60) 14 A 39 $ - $ (12.60) 39 -
I have no idea how to prove this but in playing around I found an interesting idea. How many folds using the same folding technique would it take to get back to only having 4 sides?
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Spoiler
2X4X20
Areas 8:40:80
since 160 broken down into its base is 5x25. this meant that since the ratios of B and C both contained a factor of 5, the shared side had to be a multiple of 5*2x. the rest had to be 2x where the sum of the 3 unknown powers had to sum to 5. This takes you to a very limited number of possibilities which are easy enough to just calculate.
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Spoiler
Make a cut that is along the line created by the center of the cake (if it were whole) and the center of the rectangular piece missing from the cake.
Since and straight line cut through the center of a rectangle is equal areas when you make a cut that goes through the center of both, you are taking equal area of both the cake and the missing piece so the remaining has to be equal.
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1
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Spoiler
350. Also happens to occur at 770, 1190, 1610... 350+n*420
50 groups of 7
348 left, 58 groups of 6
345 left, 69 groups of 5
340 left, 85 groups of 4
333 left, 111 groups of 3
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A little shorter then the obvious
Spoiler2+4*sqrt(2) or 7.65.
Villages numbered like a phone keypad. Road from 1->5->9. Road from 2->4. Road from 6->8. Road from 2->3. Road from 7->8.
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1
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Then try
Spoiler9,1,1,1,0
Similar logic as my previous answer but needing more than a tie to be ratified.
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Assuming a tie vote of the remaining means its ratified
SpoilerThe oldest suggest 9,1,0,2,0
Simply if the it gets down to the last 2 the youngest will get all the gold so the youngest will always vote no. The second youngest knowing this will want to make sure that it doesn't get there so will vote yes for anything 1 or more with only 3 people left. So the third youngest would ideally suggest 11,1,0 split to get the split vote and 11 for himself. This leaves the second oldest with no possible options since the middle is expecting 11 gold and the youngest is expecting 12 so nothing her can do will please a majority of bankers and give him any gold. The oldest knowing all of this just has to make sure that the second and fourth youngest are getting more then they would if the first vote wasn't ratified. knowing that going further the 4th is destined for none and the second is destined for 1, he offers 1 and 2, respectively and can keep the rest for himself.
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Spoiler
I got 32 this way. No proof it's the most
x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x -
15
1 0 1 1 0
0 1 1 1 0
1 1 1 0 0
1 1 0 1 1
0 0 0 1 1
or any 90 degree rotation of this.
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1
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edited: Misread
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A=3
B=4
C=6
D=1
E=0
F=9
More or less used inteligent guess and check knowing that some numbers had to be multiples of others already sloved for.
Bowling for equilateral triangles
in New Logic/Math Puzzles
Posted
My try at it
Consider the middle pin to be any color (since it doesn't matter) so lets say "o". Now consider all the pins adjacent to it and note that the adjacent faces of the hexagon have to have either both be the opposite color of the center or one has to be the same and the other has to be different from the center. There is only one way (really a bunch with mirroring and rotating) that this can be accomplished without creating the same color triangles
x x
o o o
x x
No consider the bottom two corners of the larger triangle. They would have to be "o" in the schematic because of the bold underlined "x"s. With both of those the top would have be an "x" because of the largest triangle possible would be the same color but this can not be because the triangle created by the top 3 pins.
Therefore this is not possible and this guy is clearly hammered since he couldn't see this.