bonanova
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Everything posted by bonanova
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Assume observer in the northern hemisphere looking southward, seeing the moon rise to the left [E] and set to his right [W] Beginning with new moon at the top, waxing to full and waning again to new. Observe the lighted side of the night moon before and after midnight. I'm interested to hear if I've sketched this correctly.
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Four nested cups fit together with no space between. The walls and bottoms are of equal thickness t. The internal heights hi and diameters di have the same ratio = hi/di for all i. The diameter d4 of the largest cup is 1 foot. The interior volumes Vi of the 3 smallest cups together equals the internal volume V4 of the largest cup: V1 + V2 + V3 = V4 What is t?
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You've had the warm-up. Now this. This may be the most difficult puzzle of its type: no starting numbers are given. [The zeros simply remind that you're beyond the last digit of the dividend.] The quotient of course has a decimal point [not indicated] but it's easy to see where it goes. The last nine digits repeat. The calculation is not shown beyond the point [2nd asterisk] where the process repeats [from first asterisk] Enjoy! ..........._________.........________ ........XXXXXXXXXXXX <- last 9 digits repeat XXXXXX /XXXXXXX ........XXXXXX ........XXXXXXX ........_XXXXXX .........XXXXXX0 .........XXXXXXX ..........XXXXXX0 <- * Same digits as below .........._XXXXXX ...........XXXXXX0 ...........XXXXXXX ............XXXXXX0 ............XXXXXXX .............XXXXXX0 .............XXXXXXX ..............XXXXXX0 ..............XXXXXXX ...................XX0000 ...................XXXXXX ...................XXXXXX0 <- *
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Gold star [and red badge of courage] for CL. This problem might be like the extra-tough SUDOKU problems that do require some trial and error. You bracketed the answer to a searchable size, and that might be the best you can do.
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It is all about energy conservation. The higher you go the higher your potential energy. When you run you gain kinetic energy. When you jump this kinetic energy is converted to potential energy. There is a limit to how fast the human body can run, and therefore how high one can jump. There is another factor. As the vaulter transforms his kinetic energy to potential energy, he raises his center of mass by an a mount v2/2g. But he raises his center of mass an additional amount by pulling downward on the pole as it nears vertical position. Think here of performers on still rings and high bar, who begin with no kinetic energy. Thus the strength of the vaulter's arms comes into play as well as his top running speed.
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Here's the other proof:
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Yeah, that's it.... Nice.
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A series of mathematical steps. Whether you program it, actually, or do it by hand. This problem you can do by hand - doesn't require the power of a computer.
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In other words, during the day you can look at the sun to know where it is. During the night, you can look at the moon to know where the sun is. Whatever information the sun gives about east and west during the day is given by the moon during the night. Let's make a rough table. From Sunrise to Noon: Sun can be seen in the EAST From Noon to sunset: Sun can be seen in the WEST From Sunset to Midnight: Moon "points" to the WEST [where the sun is, out of direct sight] From Midnight to sunrise: Moon "points" to the EAST [where the sun is, out of direct sight] This method gives an instantaneous reading of east/west. It's least precise around noon and midnight, best around sunset and sunrise. Tracking the sun and the moon as it moves across the sky also gives a reading of east and west; And it does very well around noon and midnight. But it does require a certain time interval to perform. And both methods suffer, of course, from the fact that the moon is not always visible every hour of the night. Depending on its phase and our latitude.
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Can you construct an algorithm to produce a specified-length string of consecutive integers that are not prime? For example, construct a string of 12 consecutive integers that are not prime. Then construct the longest possible string of consecutive integers that are prime!
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Both in ballpark. No definitive optimal answer - yet.
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Not quite ...
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Someone gave you a plank of wood with holes shaped as a triangle, square and circle of the indicated dimensions. Also, a 2"x2"x2" cube of wood. Can you make a single shape from that cube that will pass through each hole and completely fill it?
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APL2. Amazingly quick for prototyping/simulations. My "code" is a mix of actual code and description of code. I reproduce both of your [delta=1] results.
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The waiter wondered whether the plates stacked in his cupboard, 180 in all, would fit on the circular table without overlapping - each other or the edge of the table. That is, someone under the table would not see any part of a plate hanging over the edge. He began by centering a line of plates along a table diameter, and found that 15 plates exactly fit the criteria he had in mind. That is, the diameter of the table was exactly 15 times the diameter of the plates. A few more plates began to fit a pattern, and the waiter decided to answer his question with pencil and paper, rather than to move all the plates; and perhaps have to wash them all Forgot to mention: you are the waiter. What did you find? If 180 is not the exact number that will fit, what is that number?
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You have it. Good points, all.
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Yes! You're both on the right track exactly.
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Comments in red inside spoiler
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That's an eclipse.
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Here's a long division problem. There is a 7 in the quotient. Can you find the other numbers? .....___X7XXX XXX /XXXXXXXX .....XXXX .......XXX .......XXX .......XXXX ....... XXX .........XXXX .........XXXX .........----
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Arccos[x], or cos-1[x] is the angle theta for which cos[theta] = x. Do you want the the average value of cos-1[x] for x in some range? Not for x from 0 to pi, because cos-1[x] is not defined when |x|>1. That is, there is no angle theta for which |cos[theta]| = pi. If you want the average value of x over the region for which cos-1[x] goes from 0 to pi, that would be 0. cos-1[1] = 0 cos-1[0] = pi/2 cos-1[-1] = pi
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This puzzle was posted in 3-color form here. In discussion it was generalized to n colors. The best 3-color solutions were from octopuppy and Steve17. In the spirit of fresh debate, this thread will persist.
