superprismatic
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It is possible if you define "distance" in terms of angles instead of the usual metric.
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Once in a while, one sees a column with a mathematical bent in a newspaper. More often than not, there's something that doesn't seem quite right about it. In the Chicago Tribune column found here, a claim is made that you could have purchased 10-year bonds with a 3% rate at the beginning of 2011, sold them at the end of 2011 (when similar 10-year bonds had a 2% rate), and made a profit of around 16% on your original investment. Can anyone offer any mathematical justification for this claim?
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I think that the Professor meant that rotations which allow mirrors not to be completely contained in consecutive cells are not allowed. This makes the number of ways to place the mirrors finite, albeit very large. If my first attempt isn't acceptable, I trust Templeton will post that.
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Thanks, I made a mental miscalculation. I think that I completely understand the problem now. Time to attack it!
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You answered my question, "If the smallest mirror were to be placed within the grid block at 2B, where would the beam exit the grid (assuming no other mirrors were used)?" with "It would exit at the middle of 3A." In this case, the total path length would be 1 (4 diagonal bits each of length ¼). Is this correct?
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A little clarification, please: If the smallest mirror were to be placed within the grid block at 2B, where would the beam exit the grid (assuming no other mirrors were used)?
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Ah, Yes! I forgot about the 11 more or less thing! So, your next would be 53, whereas mine would be 42. Genius!
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x,y, and z must all be greater than 2.
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BEAL'S CONJECTURE: If Ax + By = Cz , where A, B, C, x, y and z are positive integers and x, y and z are all greater than 2, then A, B and C must have a common prime factor. EXAMPLE: 36 + 183 = 38 COOL THING: As far as I can tell, a prize of $100,000 is still unclaimed for someone who can produce either a proof or a counterexample of this conjecture. Does anyone have any ideas? I'd love to see someone from this forum win the prize!
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I neglected to add that the three beings have different skin colors. But, the smart people on the Den somehow always catch my omissions. Thanks, gadaju, for pointing out that the problem, as stated, has two possible answers. I also quite shocked that Mr. Green himself (j.green) answered the question about Mr. Green! What a coincidence! Thank you all for taking an interest in this problem.
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On the planet Phosphoron there are beings of three skin colors, red, green, and blue. Three beings, Mr. Red, Mr. Green, and Mr. Blue were discussing the beauty that this variety gives to Phosphoron. "I think it's interesting," says Mr. Red, "that none of the three of us has the skin color that our names would suggest." "Who the heck cares?" replies the fellow with green skin. What color skin does Mr. Green have?
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How many N-long binary strings are there which don't contain consecutive ones?
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Yes, you correctly understand what I meant.
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Suppose I have 2520 coins, each with a value of 1. I can easily make N piles of coins with each pile having the same value, for N=2,3,4,5,6,7,8,9,10. You may make a set of coins of any positive integer denominations you would like as long as the total value of all the coins is 2520. What is the minimum number of coins you could make this way such that you could split the coins into N equal- valued piles for each integer N from 2 to 10? For example, I can make 2269 coins which can satisfy the piles criteria as stated above: 2268 coins each having a value of 1, and one coin having a value of 252. But 2269 coins is unlikely to be minimal. By the way, 252 is the largest value a coin can be, otherwise 10 equal piles can't be made. I don't know the answer to this puzzle. I figured that I'd put it out on the Den and see what a bunch of good, imaginative puzzle-solvers can do with it!
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Can someone explain what Anza Power's code is doing? I tried, but I don't understand Java. Please explain the algorithm.
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On looking over this, I see that I still missed some. But, I'm tired. I'll fix it tomorrow.
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