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Bad news guys, I win: But a mathematician will not agree.
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i'd probably pick
- Last week
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Using the same method that BobbyGo took, my number would be a...
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A game token costs $10 to play. The pay out is $100. You can purchase multiple entries if you desire. For each entry you purchase, you must pick the lowest positive number that no one else picks. If there are ten people, including yourself, seeking to purchase tokens, what is your strategy?
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Can you post an example?
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From this screenshot, I can say that Questions that might help work toward a solution if it’s neither of those: Can you say what format the door ID is: Number of characters? All alphabetical? All numeric? Both alpha and numeric? Colors? Coordinates? Is there a reasonably high chance to expect that this should end up spelling out a phrase in plain text when we find the solution? Any previous puzzles in the game building up to this one that we should know about in order to have a sense of how we’re supposed to approach it?
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Friend gave me this puzzle. Told me not to overthink it. Im so stuck!
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K i'm posting it.
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I'm not very smart, but could you find it in your heart to post the question for me anyway?
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Hello, I did couple online IQ tests with Raven's matrices and there are some matrices where i can't figure out logic beside them and they are driving me crazy. Tests have 3 x 3 matrix and there is some logic sequence, which you have to figure out and choose the missing piece from six samples that are given to you. Do you have any idea?
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PM me for the question but only if you're smart.
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yes
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I think gavinksong is right. I sat down and did the math for the case where you're given that one focus is at (0, 0), and two points on the ellipse are at (x, 1) and (-x, 1) so the second focus must be somewhere on the Y-axis (since that math is reasonably tractable). As you move the second focus around and calculate the point at the top and/or bottom of the resulting ellipse, there are no repeats. Meaning that there is no point on the Y-axis where you could put a third point and have any ambiguity about the ellipse. My original argument was thinking in terms of a circle around the third point that contacts the curve defined by the first two points at multiple sites, but that doesn't happen if the third point is on the curve (at least in this case) -- if you think in terms of the picture from my previous post and gradually changing the value of R so you have two potential locations for the second focus along the curve which move outward along the curve as R increases, then apparently the circle around the third point grows fast enough that it "outruns" the point moving along the curve and doesn't intersect it again. If you were to move the third point farther away from the curve (when I did the math for the case above, I was covering cases where the third point sitting on the curve of potential focus locations), then the further away it gets from the curve the more distance along the curve will be covered as a circle around the third point increases in radius and the less likely you are to see any second intersection. So since it doesn't happen for points on the Y-axis, it seems unlikely to happen anywhere. But I can't definitively prove that three points will be always be enough, especially for cases where the first two given points are different distances from (0, 0) and the curve of potential spots for the second focus is not linear. (Well, there's the trivial case where all of the points including (0, 0) are co-linear so the "ellipse" is a line segment, but I'm not counting that case.)
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Proof that there exist only one answer
bonanova replied to BMAD's question in New Logic/Math Puzzles
Maybe, maybe not. -
I, as well, stand by my answer. Caveat:
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To clarify. You don't have any of these abilities: Make a line segment of equal length to another segment, or equivalently make equally-spaced pairs of points. (You don't have a pair of dividers.) Make a line segment that is parallel to another line (segment). Make an arc. (You don't have a compass.) But you do have the equivalent of a straight-edge.
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Don't see how it matters, unless ...
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Prove that x^(1/x) = x^-x has only one solution
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You and four friends are playing Russian roulette, one bullet is in a chamber of a six chamber gun. Each of you must take a shot from the gun. The chamber will only be spun once, before anyone has taken a shot. If you got to chose, which position, 1st, 2nd, 3rd, 4th, or 5th would best help your chances of survival?
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is there another?
