bonanova
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Everything posted by bonanova
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EDIT: Typo So ... would that be possible? Equivalent approach: Think of the cubes as equal in size. See if that problem is solvable, and if so, is there any extra clearance? If so, then the one with the hole can be shrunk very slightly and preserve the solution.
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Hi Supandi, Nice going. But you all know how much I love "showing your work" even tho sometimes I don't. So let me expand the puzzle. Prove it [elegantly] Do any other values of x satisfy the equation?
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No stretching or compressing, guys ... sheesh!!!
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Hi woon, I don't know how to discuss your answer without spoiling the puzzle, so ... let me promise to give you a reason later but say for now your solution won't work.
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The only presumption is a fair die and each roll gives a result.
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OK. Fair enough.
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* takes a quick glance at the puzzle title, hoping no one notices, and then confidently replies * Three.
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Can a cube be passed through a hole in a smaller cube?
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Analogies are only suggestive, to guide thinking. By achieving 1/5 of his objective we mean travel 1/5 of the distance from start to finish, not 1/5 of the remaining distance. We can take away the determinism by saying the man had walked to the city a large number of times in the past and on average it took him 5 days to arrive. If a day were 8 hours of walking, and on different days the man was well rested or not, or the weather was conducive or not to travel, then each time he made the trip his arrival time would be different from 40 hours although 40 hours was his average result. Then it is valid to assert that, on average, a day's travel achieved 1/5 of the total distance. You get what I mean. And then the equivalence of that statement with the statement that on average it takes him 1/(1/5) = 5 days is clear. However, the first argument is a rigorous proof, not that the infinite series isn't, and is simpler. Six trials, six equally likely results. On average, one result each.
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Take a variable x and raise it to the xth power: xx; call the result y. Now raise x to the y power: xy; call that result y. i.e. replace the previous value of y with the value of xy. Again, raise x to the y power: xy; and replace the previous value of y with the value of xy. Repeat this an arbitrarily large number of times, and set the result equal to 2. That creates the infinite exponential sequence x to the x to the x to the x to the x to the x .... = 2. What value of x, if any, satisfies the equation?
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Corollary: If the probability of achieving result R in one trial is p, the average number of trials necessary [expectation value] to achieve R is 1/p. By analogy. A man is walking on a road to the city named Result. After one day he achieves 1/5 of his objective. What is the expected number of days walking required for him to arrive at Result? That would be 1/(1/5) = 5 days. If 1 trial achieves a fraction p of the desired result, on average, 1/p trials are required to achieve the result. If 1 roll of the die produces a 6 with a probability 1/6, then expect 6 rolls to produce a 6.
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Wondering what all these infinite series are about ... After 6 rolls of the die, 6 numbers will appear. The average number of appearances of the numbers 1-6 is the same [for a fair die] and they total 6. If 6 quantities are equal and total 6, each of them equals 1. <_<
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No one has it yet.
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Believe it ... the rightmost number is not 8. Can you spell red herring.
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You have 5 circular disks all the same diameter and equally spaced horizontally and vertically. Make a dot in the center of the red disk. Draw a line segment through this dot that divides the 5 disks into two equal areas.
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Bingo.
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Nope. I should have said fill in the missing number. It's not the last number in a sequence: take your cues from the arrangement given.
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I know it's unlucky, but fill in the 13th number: Edit: 13th only means there are 13 numbers and 12 are already there - not that it's the last number in a sequence. ..99..45..39..36..28..21 72..27..18..21..?...13..7
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Here are three number strings: 5 8 2 7 4 9 9 4 7 8 5 5 8 7 9 which string comes next? 7 8 5 8 7 9 9 8 5 9 7 8
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I agree. My post applies only to rectangles.
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Hi lunkkun and welcome to the Den. This is an old favorite that has already been posted here. Use Search when posting puzzles you haven't created. Hope you enjoy BrainDen.
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Results so far, I believe, are: 4.5 - telethar 3.9 - telethar 3.2 - bonanova [details: cut into halves [1 cut], fifths [3 cuts], fifths [3], halves [1] -> 100 1x1 squares] 3.1 - Imran 2.995 - d3k3 3.055 - d3k3 oops 2.995 - Prime - modifying d3k3 Prime asks, can a minimum cut length be calculated? I gave that some thought, and it seems 7 cuts and 2.75m are lower bounds.
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Check your last two steps. You may have overcounted.
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Kudos to anyone who has a better answer than 3.2 meters, which is my best result. Details would be lovely to hear: what are the cuts, and what shape are the final pieces?
