jordge
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Posts posted by jordge
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I would start out with 4 Dark Stars in the shape of a polyhedron (three sided pyramid) large enough to contain the Bright star and touch each other, surrounded by another polyhedron of Dark Stars (rotated to cover the existing gaps) large enough to contain the inner polyhedron and touch each other.
Total of 8 Dark stars.(Although the scale of the problem makes my head hurt)
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483,207
Looking at a very small number of inhabitants, say 2, the number of hairs on their heads would be zero and one. Extrapolating that to increasing numbers, you find that for each number (n), the distribution of hairs is 0 to n-1. Once you get past 483,207 (to 483,208, for example) and none of the inhabitants have 483,207 hairs, and they all have different numbers of hairs, one of them must have more than 483,207 hairs which violates the fact that there are more inhabitants than hairs on any individual. -
Hexagon - Hanging it from one vertices will cause another to be at the very bottom. The remaining six vertices will be equidistant from a horizontal plane cut through the center, three above and three below. All six sides of the cube will pass through the horizontal plane, each with the same percentage of their area above or below the plane. (Three will be reversed from the other three) The shape created on the horizontal plane will have six sides with equal length and equal angles between then - Hexagon
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The remaining 2/3 can be evenly split into 400 blue and 400 pink and green, but it is also possible that there are 23 blue elephants and 777 pink and green. There is no "definite" here.
your 'observation' is only valid if those were the only colors allowed. For all we know, there is one blue, one pink and green and 798 grey elephants.
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This is a riddle that is probably best suited to being told verbally where the question mark can be inferred by the listener. In written form, the question mark defines the meaning of 'What".
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1
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Whatever the solution is, I'm sure there are more pieces than I can eat.
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Can we assume gravity as a free move? If i move locker number 1 out will locker number four safely move with gravity on top of number five?
I believe that it is an overhead view so all of the lockers are sitting on the ground in the center of the room.
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How about 10 moves to get them in the required order in the center of the room.
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Fixed amount of money.
In both cases the average price is (total cost / number of shares)
By purchasing a fixed number of shares each period, the average price is just the average of the share price since the total cost is consistently share price times number of shares. (number of shares factors out)
By purchasing a fixed dollar amount each period, the average price becomes the 'weighted' average of the cost of the shares. Since you buy more shares when the price is low and fewer shares when the price is high, the weight shifts toward the lower cost.
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Like I mentioned before, I think there are two possibilities to the switch/light combinations (determining which through the steps performed).
The three lights lit by the single switches are different from the three lights lit when all three switches are up. In which case the three 2-switch combinations illuminate a light from the single switch and a light from the triple switch combination.
1...........A
2...........B
3...........C
1,2........B,D
2,3........C,E
1,3........A,F
1,2,3.....D,E,F
The three lights lit by the single switches are the same as the three lights lit when all three switches are up. In which case the three 2-switch combinations illuminate a different pair of the unused lights.
1...........A
2...........B
3...........C
1,2........D,E
2,3........E,F
1,3........D,F
1,2,3.....A,B,C
In either case, when a light is on, a particular switch is up, regardless of what other switches may be up.
In my suggested solution, the first test indicates which possibility it is and shows one of the single lights. -
BobbyGo - I think your assumption about the light configurations is wrong. When I read the OP, I took it literally in that if a light is on, a certain switch must be up, exclusive of any other switches that may be up. This would prevent the possibility of switch 1 operating light A individually but then the switch 2 and 3 combination also illuminating Light A.
OP - Clarification?
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I think it can be done with 3 tests.
My understanding of the rules is that there can be two possibilities for combinations.
- Possibility 1: Identified in that the three lights on the single switch combinations are different than the three lights on the three switch combination. The two light combinations comprise of one light from a single switch and one light from the three switch combination.
- Possibility 2: Identified in that the three lights on the single switch combinations are the same as the three lights on the three switch combination. The two light combinations comprise of different pairs of the three remaining lights.
Step 1 - Turn on switches 1, 2, and 3. Wait. Turn off switches 2 and 3. Check the lights.
Step 2 - Turn on switch 2. Wait. Turn off switch 1. Check the lights.
Step 3 - turn on switch 3. Wait. Turn off switch 2. Check the lights.
I'll leave it as a further exercise to work through what information each test provides. -
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How would your calendar show Wed. June 19, 2013?
A better question is, "How does the calendar show Wed, 19 Aug"?
Use the 'M' from the first dice - turn it over to make the 'W'.
Use the '6' and '1' from the next two dice - turn over the '6' to make the '9'.
Use the 'g', 'a' and 'n' from the next three dice - turn over the 'n' to make the 'u'.
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M T Th F Sa Su
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The shortest time - 6408 econds based on 1 second for each square foot mowed and 158 turns.
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176 Feet
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mn-1. You can have it either a. Breaking the m rows first (m-1 breaks) and then breaking each of the m sticks into n pieces (total of m(n-1). In all mn-1 b. Breaking the n columns first (n-1 breaks) and then breaking each of the n sticks into m pieces (total of n(m-1). In all mn-1
There are actually more ways to break it up, not just do all m and then do all n. You could pick up any joined piece and break it in any order. But looking at your example for m first:
Break all the m rows: (m-1) breaks
Break all the m sticks into n pieces: m(n-1) breaks or (mn-m) breaks
Add all the breaks: (m-1) + (mn-m) = mn-1
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Every break increases the number of pieces by one by taking a larger piece and turning it into two smaller pieces. You start out with one large piece and continue breaking until all pieces are as small as allowed. If the total number of small pieces is n*m then the number of breaks to get that many is always (n*m)-1.
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Nested Figures in a Pentagon
in New Logic/Math Puzzles
Posted
When you said the blue diamond, did you mean the blue star? If so, it would have a larger perimeter than the red pentagon. But that is all I can determine.