k-man
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Everything posted by k-man
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Sorry, swongy, but your solution is not complete. I didn't go through the whole thing since I found a bad assumption right in the beginning: If both 2 and 3 are heavy (or light), or both 1 and 4 are heavy (or light) you will not discover them using your method. Since no one has solved this yet I will work on the solution and post it soon.
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Here is an additional challenge: Prove that for any number k (k>2) of points placed on the flat surface the maximum number of triangles that can be formed without crossing the lines is always (k-2)2 The proof is trivial for k=3 and k=4, but can you provide the general proof for any k>4?
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The autopsy can show if the cause of death was drowning or not. From the riddle it sounds like both victims were chocked before they were placed in the water.
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If the lines were allowed to cross then this answer would be correct, but the lines cannot cross.
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Mulikar got it! Good job. BobbyGo was pretty close and even posted a picture of how to arrange the dots, but lost one triangle by aligning the bottom 3 dots horizontally.
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How many distinct triangles can you make by placing 10 points on a flat surface and connecting them with line segments? Line segments cannot cross.
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Duplicate post. Answer this thread instead
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Read the wikipedia page that Anza referred to in post #2. It has a wealth of info about the magic squares including the techniques of producing them.
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Sorry, but you're not quite there yet. You didn't detect 3 cases of 2 fakes: [2,6],[3,4] and [3,5]. Generally speaking, when you have 2 weighings left and 12 possibilities, you should rethink your approach. 2 weighings will produce 9 distinct results, so you cannot distinguish between 12 different possible scenarios. As a side note, I think you simplified the problem by assuming that if there are 2 fakes then one is heavy and one is light. I don't interpret the problem this way. In my interpretation both fakes can be heavy or both can be light. In this interpretation we are faced with 73 possible scenarios. 4 weighings are sufficient to not only identify the fakes, but also to state for each fake whether it's heavy or light. In your interpretation you only have 43 scenarios, which still requires 4 weighings to determine the fakes and their relative weight, althogh it's MUCH easier. However, if we only need to identify the fakes without knowing whether they are light or heavy we MIGHT be able to do it in only 3 weighings.
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That's right. I stated that assumption in my answer. We both make the same assumption basically treating the plant as a renewing source of food with constant daily increment.
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Welcome to the Den, rparry. Please use spoilers when posting the answers. Your equations are correct, buy you made a mistake in your calculations. Solving these equations results in a=Uo/80 and b=Uo/1600.
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njBLOB, Welcome to the Den. Please start a new topic for every new riddle/problem. It gets messy when multiple unrelated riddles are posted in the same topic.
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I solved this yesterday too, but had to leave, so didn't have time to post my solution. It's slightly different from the one posted by plainglazed, so not to waste good 3 hours of work, here goes... I don't know if these nested spoilers was a good idea. It seemed like a good idea when I started, but now it seems to make it harder to read rather then easier. Oh, well...
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We continue these iterations until we reach Y=1, which is the object of the puzzle. The total cost depends on the pairs chosen and answers given, but the worst case scenario is when each iteration reduces the number of villagers to n or n+1 (whichever is odd). For 99 initial villagers the maximum cost of each iteration is: 1. $49 (49 villagers left) 2. $24 (25 left) 3. $12 (13 left) 4. $6 (7 left) 5. $3 (3 left) 6. $1 (1 honest villager left) The total cost is $95.
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so true!
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That's exactly what I was doing too, but I guess i didn't spend enough time and didn't have enough patience checking the permutations.
