bonanova
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This question is adapted from a [u.S.] national math competition question at the High School level. A fair coin has the letters a and B on its two sides. A fair die has two sides labeled D, two labeled e and two labeled i. Another fair coin has the letters n and r on its two sides. The coins and die are each tossed three times. Find the probability that the nine letters so obtained can be used to spell BrainDen.
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Hats on a death row!! One of my favorites puzzles!
bonanova replied to roolstar's question in New Logic/Math Puzzles
OP states that ... Of course you will not be able to see the color of your own hat -
You live in Killville. Can you stay alive?
bonanova replied to bonanova's question in New Logic/Math Puzzles
That's after one encounter. How about after unlimited encounters? -
I was thinking part 3 is equivalent to the graph theoretical question The edges of a nondirected graph are colored with m colors. How many nodes must it have to ensure a monochrome loop of size x. But now I see two discrepancies. A colored edge assumes its nodes share a color [emotion], a condition not imposed by OP. Only requiring a loop is not enough - you need all the edges [star shaped] to be monochrome. Discrepancy [1] seems to rule out a graph analogy.
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Hi Freaky, and welcome [as a member now] to the Den! You're spot on with your comment. The correct result [they all weigh the same] was reached [the main question of the OP], even tho the weight in some calculations was off by a factor 23. Looking forward to seeing some of your favs posted here.
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When I posted, I was convinced that 4 would not benefit from hex packing, but I hadn't verified that 5 would. Else I would have included only two sizes: either 1 and 5, or 4 and 5. Note that seven layers would have to fit: 25+16+25+16+25+16 = 123 compared with 5 cubic packed layers - 125.
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Excellent. With that assumption, which certainly holds for the examples in the OP, no math is needed.
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I agree: there is not enough information to deduce the official answer.
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Four identical cubical boxes, with interior dimensions of a cm on a side, and weighing b grams each, were packed with wooden spheres of density of d gram/cm3 Box 1 contained 1 single sphere; the largest one that would fit inside the box. Box 2 contained 8 identical spheres; the largest that would fit inside the box. Box 3 contained 27 identical spheres; again the largest that would fit. Box 4 contained 64 identical spheres; again the largest possible. Which loaded box [box plus contents] is heaviest, and what is its weight?
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This is a classic riddle, perhaps most famously cast by Martin Gardner as the "Unexpected Hanging". A prisoner was told he would be hanged nest week, on a day that could not be anticipated. His lawyer made the above arguments to show the execution could not therefore take place at all. He was hanged the following Wednesday!
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Six moves is minimum - nice. That page is new to me - did you construct it?
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Placing the mythical crown of the Grand PooBah ceremoniously on Grayven's brow. Congratulations.... OP gave you all an official answer form. While you weren't given the puzzles themselves, you saw answers for 10 unique scores. By solving in effect an eleventh puzzle, you were able to determine the 10 correct answers. That done, you could submit a perfect answer form and claim the prize. preflop came close: he gave what his answers "would have" been; Grayven actually submitted them. Nice going all.
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Line 2.
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Ten Brain Denizens had a shared dream of competing for Grand PooBah Puzzle Master of all time. They all achieved different scores on a set of ten 10-point puzzles: five multiple-choice [A, B, C, D, E] and five True-False. Just as they finished, you joined the dream and persuaded the proctors to let you compete. You were given an official answer form, but as a handicap for arriving late you were not permitted to read the puzzles. Who won the Grand Prize? Oh yah, here are the others' answers: Denizen........ 1 2 3 4 5 | 6 7 8 9 10 ====================================== Cherry Lane.... C C B D D | T F F F F d3k3........... A C D E A | F T F F F HoustonHokie... A B E C B | T F T T T Izzy........... C D A E B | T F F F F nobody......... C C E D A | F T F F F Prime.......... C E A C E | F F T T T Prof. Templeton A B A C E | T F F T T reaymond....... B B D C A | T F F F F Shakeepuddn.... E E C C B | T T T T T woon........... A B C C D | F F F T T If your name appears in the list, not to worry; consider this a second chance. Higher score counts.
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You have 16 coins in a 4x4 array, numbered 1-16: 1-4 is top row, etc. When you touch a coin, it flips, along with the other coins in its row and column; 7 coins in all flip. The objective is to make all the coins show tails in the fewest possible moves. For graphical clarity, O is heads, + is tails. Here's an easy example: coins 4, 7, 10 and 13 are heads. + + + O + + O + + O + + O + + + Touching coin 4 produces this array: O O O + + + O O + O + O O + + O It's easy to verify that touching coins 7, 10 and 13, in any order, then completes the task of getting all tails [+]'s. The solution thus would be written 4 7 10 13. Or 4 10 13 7. Order of moves does not matter. Flip all sixteen coins to tails [+] in the fewest moves possible: Puzzle 1 [easy] O O O O O + O O O O + O O O O O Puzzle 2 [harder] O O O + O O O O O + O O O O O O Have fun.
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Misread the OP, thought the liking was a quality like colored socks.
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Hi Liz, and welcome to the Den.
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The OP says saw, discard, and construct, in that order. But you have a point that it does not specifically prohibit further sawing. This is the biggest size possible if more sawing is permitted....
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Agree with voltage, prime and d3k3. The statements that do it most simply for me are Having only one excess coin, May cannot throw more heads and more tails. Therefore she will throw more heads or more tails. I was going to add "with equal probability" to statement 2. But the terms were to not say that word. So unsaid, but still true.