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Because the book of logic puzzles I have used twice went missing, here are the links to the first (https://forums.somethingawful.com/showthread.php?threadid=3851344) and second (https://forums.somethingawful.com/showthread.php?threadid=3870510) threads I have put these puzzles in. Our last puzzle was #39, solved correctly by Princey. Since this section (Matching Wits) is a little less interesting, I will put puzzles #40-42 all in a row so they can be solved all at once. Puzzle #40: Royal Dinner To enlist the help of the other kingdoms, the king talked to the queen about inviting neighboring royalty to dinner. This put the queen into a royal snit. Theirs was not a very wealthy kingdom and the royal dinnerware was in a disgraceful condition. Apart from ordinary dishes for everyday use, all that the royal pantry contained were a few dinner plates of three different patterns: 1. five silver ones with birds 2. six crystal with seashells 3. seven gold with the royal crest They were all stored in disarray on a very dark top shelf of the royal pantry. Only those would be suitable for entertaining other royalty. If the queen didn't want to climb up to the top shelf twice, how many dinner plates would she have to take down to be sure she had matching dinner plates for herself, her royal spouse, and for the neighboring king and queen? Puzzle #41: Anti-Ogre Potions The king had his doubts about his son's fighting skills, and so he sent his two eldest to the court magician for potions to help fight the ogre. The magician kept his magic hidden, mindful of the danger of his potent potions falling into the wrong hands. In a secret but inconvenient compartment in his laboratory, he hoarded: 1. four ogre-fighters 2. three dragon-destroyers 3. two evil-wizard-vanquishers How many potions did he have to reach for in order to make sure that he could give an ogre-fighter to each of the king's two sons? Puzzle #42: Seven-League Boots Meanwhile, back at the castle, the ogre found that the boots he had picked at random from his dark storeroom were all six-league boots. He threw them back. He needed seven-league boots so that he could cover more territory. If in that dark storeroom he had four six-league boots and eight seven-league boots, how many boots did he have to pull out to make sure he had a pair of seven-league boots?
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Nov 20, 2020 21:03
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Puzzle #40: Royal Dinner ten plates; assume the queen grabs three of each of the three options, the tenth plate has to be the fourth of one of three sets Puzzle #41: Anti-Ogre Potions seven; if the magician pulls all three dragon potions and both wizard potions, he'll need two more to get ogre-fighters. Puzzle #42: Seven-League Boots six; if he pulls out all four six-league boots to start with he'd need to get two more to guarantee two seven-leaguers.
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Nov 20, 2020 21:19
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Puzzle #40: Royal Dinner 10 Puzzle #41: Anti-Ogre Potions 7 Puzzle #42: Seven-League Boots 6 if boots he counts boots individually, 5 if boots refers to pairs of boots
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Nov 20, 2020 22:05
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Puzzle #40: Royal Dinner 10 plates. Assume that the queen gets the worst luck possible and grabs 3 of each kind of plate. The 10th must make 4 matching plates. Puzzle #41: Anti-Ogre Potions 7 potions - again, same principle. Assume the magician has the worst luck and grabs all 5 other potions first - then if he grabs 2 more times, he must have gotten at least 2 anti-ogre potions. Puzzle #42: Seven-League Boots There's three possible answers to this puzzle. If boots are interchangeable on each foot and the ogre picks boots one at a time, six boots is enough. If there are four and eight pairs of boots in the storeroom and the ogre grabs out pairs, five pairs is enough. If there are two left six-league boots, two right six-league boots, four left seven-league boots, four right seven-league boots, and the ogre picks boots individually, the ogre must pull out 9 boots to ensure a pair of seven-league boots. Y'know, maybe I should LP The Puzzling Adventures of Dr. Ecco someday...
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Nov 20, 2020 22:34
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Everyone got the first two correct, and the third's correct answer is six. The next section is In The Ogre's Dungeon, and while the last one is a bit more complex, the first four are simple enough that I'll bundle them together. Puzzle #43: In The Forest The king's only children, Able, Benjamin, and Paula went into the forest with their friend, the elderly Sir Kay. They wanted to try their skill with their bows and arrows. Each of them started with the same number of arrows. When all the arrows had been shot, it was discovered that: 1. Sir Kay brought down more game than Princess Paula. 2. Prince Benjamin capture more than Sir Kay. 3. Princess Paula's arrows went truer than Prince Abel's. Who was the best marksman that day? Puzzle #44: Captured! Happy at the hunt, the king's children became careless and less watchful than usual. A passing ogre easily captured them and Sir Kay and took them back to his dungeon. He placed them in four cells in a row. The cell in which Prince Abel was held prisoner was next to Prince Benjamin's. But Prince Abel was not next to Princess Paula. If Princess Paula's cell was not next to Sir Kay, whose cell was? Puzzle #45: The King's Heir The ogre's prisoners spent a sleepless night in their dungeon cells wondering what fate awaited them. The next morning, the ogre approached the king's sons. "Which one of you is the king's heir?" he demanded. "I'm Abel, the king's eldest." said the prince with black hair. "I'm Benjamin, the king's second son." said the one with red hair. If at least one of them lied, who lied? Puzzle #46: The Ogre's Boast "I've devoured more than one hundred humans." the ogre boasted. "Surely, it must be fewer than one hundred," said Sir Kay. "Well, I suppose it was at least one," said Abel. If only one spoke the truth, how many humans did the ogre actually devour?
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Nov 20, 2020 22:47
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43) Benjamin is the only one who is never worse then any other 44) Abel's cell is next to Sir Kay's and Paula's cell is next to Benjamin's, the question is a bit unclear. Cells go: AB, then to the left is Not P which means either K or nothing, [K_]AB[KP][KP] is possible with KABP being the only option without K next to P. and 45) Abel If B has lied then A must have lied also, but A can have lied and B told the truth if their sister is the eldest. 46) If it is 0 then Kay spoke the truth. If it is exactly 100 then Abel spoke the truth.
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Nov 20, 2020 23:17
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44: We're given that Abel is next to Benjamin, and that Paula is not next to Abel OR Kay. Since Paula must be adjacent to someone, that must be Benjamin. Therefore, Benjamin is between Abel and Paula, and Kay is on one of the ends. That can't be the end next to Paula, so Kay is next to Abel. It goes Kay, Abel, Benjamin, Paula or the reverse of that.
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Nov 21, 2020 00:09
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43: Benjamin > Kay, Kay > Paula, and Paula > Able. Hence, Benjamin > Kay > Paula > Able, and Benjamin is the best marksman. There's a name for this property where A > B, B > C then A > C, but I forget what it's called. 44: If Paula is next to neither Able nor Kay, then she must be next to Benjamin. Benjamin is thus next to Able and Paula, leaving Able as the only one who could possibly be next to Kay, since we've already ruled out Paula being next to Kay, and having Benjamin next to Kay would result in Benjamin having 3 adjacent cells. 45: If red is lying, then he's the king's first son, which means black is lying as well. If black is lying, then he could still be the prince's first or second son, the princess could just be older. They are both lying. 46: Only of one of n >= 1, n >100, and < 100 are true. The only number that satisfies these conditions is 100. The ogre has eaten exactly 100 humans.
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Nov 21, 2020 01:30
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Puzzle #43: In The Forest Prince Benjamin. Puzzle #44: Captured! Prince Benjamin again. Easy process of elimination. Puzzle #45: The King's Heir If "Abel" is lying, then he's Benjamin. And "Benjamin" is therefore lying about who he is. If "Abel" isn't lying then "Benjamin" can't be either, so that's impossible. So they're both lying. Puzzle #46: The Ogre's Boast The ogre hasn't eaten anyone! Its boast and Abel's guess is a lie. Kay is right in saying it's under 100.
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Nov 21, 2020 02:54
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Correct on 43 for all, 44 is Abel, correct on all for 45, and surprisingly, 46 has two equally valid answers: 0 or 100 exactly. Puzzle #47: Heads for Hats Keeping prisoners was much less entertaining than the ogre thought it would be. He decided to have some fun. The ogre brought in a box with five hats, two red and three white. Then he blindfolded his three young prisoners and placed a hat on each head. "Each one of you must guess the color of the hat on your own head-without using a mirror," the ogre said in his meanest voice. "I'll take off your blindfolds one by one and let you try. If not one of you guesses correctly, all of you will die." Abel, the oldest, was used to taking charge. "Don't worry," he said, "I shall save us," and he bid the ogre take off his blindfold first. He examined the hats his brother and sister were wearing and then admitted that he didn't know what color hat he was wearing. Benjamin, the second oldest, insisted that he be given the next chance. He, too, was sure that he could save his brother, his sister, and himself. But after his blindfold was removed, he too had to admit that he did not know the color of the hat he was wearing. Then Princess Paula said, "I don't need you to take off my blindfold. I can tell you what color hat is on my head." Did the three go free? What color hat was Paula wearing?
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Nov 21, 2020 03:37
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47: If any one of them sees the other two wear a red hat, then they know they are wearing a white hat. Thus, we immediately disregard the scenario where Paula and Benjamin or Paula and Abel are both wearing red hats. Let us consider the scenario where only Paula is wearing a red hat: Abel sees one red and one white hat, and there is a 2/3rds chance he is wearing a white hat and a 1/3rds chance he is wearing a red hat. Benjamin, seeing Paula wear a red hat now knows that he is not wearing a red hat, since if he was then Abel would have called out that he was wearing the white hat, and thus Benjamin calls out he is wearing a white hat. This is not the solution. Paula therefore must have been wearing a white hat, as the only scenario wherein Paula is wearing a red hat is not possible. There are now four possible scenarios: Benjamin red, Abel red, both red, or both white. Both red: Abel sees one red and one white. Benjamin sees one red and one white, and knows that Abel did not see two red hats. Paula knows only that neither of them saw two red. Benjamin red: Abel sees one red and one white. Benjamin sees two white, and knows that Abel did not see two red hats. Paula knows only that neither of them saw two red. Abel red: Abel sees two white. Benjamin sees one red and one white, and knows that Abel did not see two red hats. Paula knows only that neither of them saw two red. Both white: Abel sees two white. Benjamin sees two white, and knows that Abel did not see two red hats. Paula knows only that neither of them saw two red. There are no solutions where Paula knows she is wearing the white hat just from neither Benjamin nor Abel not knowing what hat they're wearing. The three lose their heads. 
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Nov 21, 2020 06:25
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47: I think they all survive because everyone guesses that they have a white hat on. With 3 white hats and 2 red hats, 1 of the 3 siblings must be wearing a white hat. Nobody actually needed to take off their blindfold. In the story, both Abel and Benjamin must have guessed wrong. If they both opted to guess white incorrectly, then Paula guesses white correctly.
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Nov 21, 2020 09:40
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#47 : Abel can only guess his color correctly if he sees two red hats. Therefore he either sees two white, or one red and one white. Benjamin does not see two red either, and is similarly unable to guess his color. But if he were to see a red hat on Paula, he would know for sure that Abel is seeing a white hat on himself, since Abel cannot be seeing two red hats. Therefore he must be seeing a white hat on Paula. Paula is wearing white, and we don't know what the others are wearing.
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Nov 21, 2020 13:58
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Kangra has got it correct! And now, we move onto a new section. Break out your magic carpets, because in this section we deal with The Genie's Revenge. Puzzle #48: Hidden Gold The clumsy apprentice of a wealthy Arabian merchant uncorked the char in which a genie had been imprisoned for many years. Free at last, the genie looked about the Arab's shop to see what mischief he could make. He could, of course, have destroyed the merchant's shop or even killed the merchant, but he quickly realized that the merchant valued his money much more than his life! Seizing the merchant's gold, he hid it at the bottom of a huge earthen olive jar. Then he brought in eight identical olive jars and placed three-pound weights in them. Last, he filled all the jars with olives and sealed them securely. When the merchant became distraught at his loss, the genie revealed what he had done and agreed to give the merchant back his wealth if he could guess which jar held the gold. The genie would not let him open any of the jars. He could only weigh them. The catch? He could only use the scale three times. The merchant owned a balance scale with pans on each side. How did he identify the jar with the gold?
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Nov 21, 2020 16:51
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If the gold weighs 3 pounds exactly the merchant is SOL. Or if the djin had more then 7 of the weights. If he knows if the gold lighter or heavier then the weights he can do a binary partition search. If he doesn't know if his gold is heavier or lighter things get complicated. 1)Weigh 3 coins on each side. a)If they are balanced all six are empty. Weigh each of the two left separately against empty ones to find the one that tilts. b) If the right is heavy walk to the other side, so that the left is heavy. c) Have the left side be heavy: Move 2 pots from left to right, 2 from right to outside and the two from outside to left. 2) Weigh again. a) If they are balanced one of the outside ones is gold and it is lighter, weigh them against each other. b) If left is heavy again, the stationary pots have the gold. Weigh one of them against a known pot and it is the gold if it tilts. c) If right is heavy one of pots moved from left to right has the gold and is heavy. Weigh them against each other and the heavier one has the gold. e: oops I just noticed that there are 9 and not 8 total pots. So: 1)Weigh 3 coins on each side. a)If they are balanced all six are empty. Weigh two paring of the separate ones: If they are both even the last pot is full. If they are both uneven the shared pot is full. After step 2c) discard the last outside one, it is empty. VictualSquid fucked around with this message at 22:52 on Nov 21, 2020 |
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Nov 21, 2020 17:42
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48: There's a total of 9 jars. 1 is the target with the gold. It has to be assumed that the gold weighs either more or less than 3 pounds. However, we can't assume the merchant remembers the exact weight of their gold. Measurement 1: Take 2 of the jars and weigh them against another 2 jars. 1a: Scales are even. 1b: One side is heavier. Measurement 2a: If the scales are even in 1a, then we know we have 4 olive jars, and 5 unknowns. Measure 3 olive jars against 3 of the 5 unknown jars. 2aa: Scales are even. 2ab: One side is heavier. Measurement 3aa: If the scales are even in 2aa, then we have 7 olive jars and 2 unknowns. Measure 1 olive jar against 1 of the 2 unknown jars. 3aaa: Scales are even. -> The unmeasured unknown jar has the gold. 3aab: One side is heavier. -> The measured unknown jar has the gold. Measurement 3ab: If a side was heavier in 2ab, then the merchant should be able to conclude that 1 of the 3 unknown jars has the gold and whether the gold jar should be lighter or heavier than an olive jar. Of the 3 unknown jars, measure 1 against the other. 3aba: Scales are even. -> The unmeasured unknown jar has the gold. 3abb: One side is heavier. -> Based on whether the gold jar is heavier or lighter, the merchant knows which of the 2 measured unknown jars has the gold. Measurement 2b: From 1b, there are 4 suspicious jars. The unmeasured 5 jars are olive jars. This situation (4 unknown, 5 olive) is strictly easier than the situation in 2a (5 unknown, 4 olive), so the merchant can use a similar measurement process to figure out where the gold is.
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Nov 21, 2020 21:23
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VictualSquid got it right with the 'divide into groups of 3' tactic. Puzzle 49: Baskets and Baskets The genie wasn't through with his tricks. In the merchant's warehouse were twelve sealed baskets of grain, one of which was fodder for pigs. The genie stealthily removed the labels and rearranged the baskets so that it was impossible to tell which contained pig fodder. The merchant didn't discover the situation until a customer arrived to buy four baskets of grain. The important customer was in a hurry. If the pig fodder weighted a bit more than the other grain, how could the merchant, in one weighing avoid the pig fodder and make sure he was selling fine grain? Puzzle 50: Wanted--- Pig Food The merchant's next customer was a farmer whose storehouse was empty. He needed food for animals. How many weighings did it take for the merchant to find the basket of heavier fodder among the remaining baskets?
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Nov 24, 2020 07:43
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Puzzle 49: Baskets and Baskets Weigh four baskets against four other baskets at once. If one side goes down, that side contains the fodder; sell the four baskets that are on the light side of the scale. If neither goes down, an unweighed baskets contains fodder; sell any four weighed baskets. Puzzle 50: Wanted--- Pig Food We have a head start on this one because we already know of eight baskets that do not contain pig food - we sold four, and we know four more by process of elimination. So we're left with four uncertain baskets. Of those four baskets, weigh a random one against another random one. If one is heavier, sell that (1 weighing). Otherwise, set them both aside (they're both fine grain) and weigh the remaining two baskets at once - the heavier one has the pig fodder. This solves it in 2 weighings, or 1 if the merchant is lucky. Quackles fucked around with this message at 07:53 on Nov 24, 2020 |
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Nov 24, 2020 07:51
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49) He can divide them into 3 piles of 4. Then weigh two piles, the fodder is on the heavier side or the leftovers if it is even. Or he can divide it into 2 piles of 6. The fodder is on the heavier side and he will have two known good pots leftover. 50) That is the standard ternary search: Divide into 3 equal groups, round up to 3^m pots with the missing pots in the outside group. Weigh two groups and take the heavier or the outside in the even case. Then repeat with the winner group until you have sufficiently narrowed it down. This will take log_3 n weightings in this case 2.
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Nov 24, 2020 14:25
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Puzzle 49: Weigh three baskets against three other baskets. If they are evenly balanced, the merchant can sell any four of them, and if not, one of the three heavier baskets have the fodder and the rest are fine grain (since only one basket has fodder in it). Puzzle 50: If the weighing from Puzzle 49 didn't come out even, only one weighing is needed: pick any two of the three possible fodder baskets, and either the heavier basket in that weighing is the fodder or, if the weighing is even, the unweighed basket is the fodder basket. If it came out even, two weighings are needed: first to find three of the six unweighed baskets that could be the fodder, and the second as in the earlier instance.
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Nov 24, 2020 14:33
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49: Weigh four baskets against four other baskets, leaving four baskets aside. No matter what, that gives you eight known-good baskets and one group of four that includes the fodder basket. If the balance was even, the unweighed group has the fodder. If the balance was uneven, the heavier group had the fodder. Either way, sell the customer four of the eight known-good baskets. 50: We now have four known-good baskets left over from puzzle 49, and four suspicious baskets. Worst case you'll need two more weighings, but there's a way to give you a 50% chance of needing only one weighing. Take three of the suspicious baskets and put two of them on the left pan, and one of them on the right pan. Then put two known-good baskets on the right and one known-good on the left. Now if the left side goes down, then one of the two suspicious baskets had the fodder and you'll need a second weighing to determine which. But if the right side goes down, you've identified the fodder in a single weighing. And if the pans are even, then the single leftover suspicious basket has the fodder. So basically, the chance of needing a second weighing is equal to the chance that the fodder was among the two baskets you picked for the left side. I won't spell out all the combinations, but if you work it all out it comes to a 50% chance that you put the fodder on the left pan. So this is the most efficient method, with a 50% chance of needing just one more weighing, and a 50% chance of needing two. (Most of the other solutions so far have had a higher chance, generally 75%, of needing two weighings.) And since I'm catching up... Aesculus, here's what was wrong with your answer to 47, in case you're interested: You correctly analysed that Paula had to be wearing a white hat, but your later analysis that "Paula only knows that neither of them saw two red" doesn't give Paula (and Benjamin) credit for being able to do the same analysis. After Abel can't guess, Paula knows that Abel didn't see two red hats — and she also knows that Benjamin knows that Abel didn't see two red hats. After Benjamin can't guess, Paula knows that Benjamin didn't see two red and that he didn't see a red hat on her own head. Because if Benjamin had seen a red hat on Paula's head, he would have known that since Abel didn't see two red hats his own hat can't be red (otherwise Abel would have answered "white") and he would have answered the question. So the fact that Benjamin was unable to figure out the answer tells Paula that her hat must be white, and so she can safely guess white and they all live. Tax Refund fucked around with this message at 10:34 on Nov 25, 2020 |
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Nov 25, 2020 10:23
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All correct for 49, and the answer for 50 is 2 weighings. Puzzle 51: Lead Weight Thwarted by the merchant's ingenuity, the genie spirited away the merchant's scale and weights. But the merchant made a scale by balancing two empty baskets on either end of a long pole. Then he got a piece of lead weighing exactly fifteen ounces. He cut the bar into four pieces so that he could weigh objects from one to fifteen ounces. What were the weights of the four pieces he cut? Puzzle 52: Heavier Stakes The merchant still had a problem weighing the heavier merchandise in the bazaar. He bought a forty-pound bar of lead. If he cut the bar into four pieces so that he could weigh items from one pound to forty pounds, what would each piece have to weigh?
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Nov 26, 2020 05:13
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Puzzle 51: Lead Weight 1, 2, 4, and 8 ounces.
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Nov 26, 2020 05:51
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Puzzle 51: 1, 2, 4, and 8 ounces. Puzzle 52: 1, 3, 9, 27 pounds.
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Nov 26, 2020 06:23
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Oh, I hadn't realized he could put weights in both pans for puzzle 52.
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Nov 26, 2020 06:29
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Quackles posted:Oh, I hadn't realized he could put weights in both pans for puzzle 52. Really, I only came up with the answer by idly wondering if trinary would be used there and realizing that that approach did add up to 40 pounds. I don't really know the mechanics of it offhand! e: Thinking on it, you're right, that's how it has to work for a precise determination. But hey, if the puzzle doesn't say it's explicitly against the rules to do it... Ignatius M. Meen fucked around with this message at 10:48 on Nov 26, 2020 |
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Nov 26, 2020 10:40
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Correct, Ignatius! Puzzle 53: Weighty Matters To weigh a forty-pound object with four weights-one, three, nine and twenty-seven pounds-the merchant placed all of the weights on one side fo the scale and the object on the other side. But how did he weight objects weighing a) five pounds? b) fourteen pounds? c) twenty-seven pounds? d) twenty-five pounds? yes, that is the riddle. It's dumb. I'll throw in the last one of the section as well. Puzzle 54: Gold and Silver Coins The genie was still making his way through the merchant's shop, messing up whatever he could. The merchant had ten sacks, each containing ten coins. In one sad the coins were silver, in the others were gold. The genie slyly coated all the coins bright red and put them back in their original sacks. The merchant knew that a gold coin weighed ten grams and that a silver coin weighed a gram less. If he used a regular scale, how could he determine in one weighing which sack was not gold?
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Nov 27, 2020 07:40
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Puzzle 51: Assuming he doesn't need fractions of an ounce, the simplest solution (and the obviously intended one) is 1, 2, 4 and 8. But using the same technique in problem 52's solution, there would be other possible answers like 3/8 oz, 1 1/8 oz, 3 3/8 oz, and 10 1/8 oz. That would allow him to weigh any amount from 3/8 oz up to 15 oz, with a precision of 3/8 oz, by combinations of weights in both baskets. This doesn't yield integer results — he couldn't weigh a 1-oz object exactly, for example — but in the case of a merchant, being able to say "this object is heavier than 6/8 oz and lighter than 9/8 oz" would be more useful than saying "this object weighs more than 1 oz but less than 2 oz" (because merchants usually deal in things like fruit or spices, whose weights rarely come to exact multiples of 1 oz). Puzzle 52: 1, 3, 9, and 27. To weigh a 2-oz object, put the 1-oz weight with it and the 3-oz weight on the other side. To weigh a 6-oz object, put the 3-oz weight with it and the 9-oz on the other side. To weigh an 11-oz object, put the 1-oz weight with it and the 3 and 9 on the other side. And so on; by different combinations of adding and subtracting the numbers 1, 3, 9 and 27, you can get any integer from 1 to 40.
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Nov 27, 2020 07:41
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Puzzle 53: Yes, this is dumb; if you solved puzzle 52 you've already solved this one. 5: put the 3 and 1 with it and put the 9 on the other side. 14: put the 9, 3 and 1 with it, totalling 27, and put the 27 on the other side. 27: Seriously? Is this supposed to be hard? 25: put the 3 with it, and put the 27 and the 1 on the other side so that both sides should weigh 28. Puzzle 54: Number the sacks 1-10 and take one coin from sack 1, two coins from sack 2, and so on up to all ten coins from sack 10. Total of 55 coins. If they were all gold they would weigh 550 grams. If they weigh 540 grams, then ten of them were silver, so sack 10 had the silver coins. If they weigh 541 grams, nine of them were silver so sack 9 had the silver coins, and so on down to 549 grams (in which case sack 1 had the silver coins).
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Nov 27, 2020 07:47
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Puzzle 53: a) Place the nine pound weight on one side, and the object plus the one and three pound weights on the other. b) Place the twenty-seven pound weight on one side, and the nine pound, three pound, and one pound weights plus the object on the other. c) Place the twenty-seven pound weight on one side, and the object on the other.  Especially strange and dumb even for this question...d) Place the twenty-seven pound weight and the one pound weight on one side, and the object plus the three pound weight on the other. Puzzle 54: Number the sacks 1 to 10. Take one coin from sack 1, two coins from sack 2, etc. up to ten coins from sack 10, and weigh all 55 coins on the scale. The total weight will be between 540 to 549 grams; subtract this weight from 550 grams and the result is the number of the sack which has silver coins in it.
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Nov 27, 2020 08:31
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Correct! And now, we continue the story of this genie and merchant with the next section, Genie Devilment. Puzzle 55: The Brothers Four Angered at his failure to trick Abou, the merchant, the genie transformed him and his three brothers into animals. He turned one into a pig, one into a donkey, one into a camel, and one into a goat. 1. Ahmed didn't become a pig, and he wasn't a goat. 2. Sharif wasn't a camel, and he wasn't a pig. 3. If Ahmed was not a camel, Omar was not a pig. 4. Abou didn't become a goat, and he was not a pig. 5. Omar was not a goat nor was he a camel. What did each of the brothers become?
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Nov 28, 2020 03:47
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Puzzle 55: The Brothers Four OK, Ahmed can only be a camel or a donkey. Sharif can only be a goat or a donkey. Abou can only be a camel or a donkey. Omar can only be a pig or a donkey. Notably, Ahmed and Abou are both tied in contention for camel and donkey, so Sharif and Omar can't be a donkey. So, Sharif must have been turned into a goat, and Omar into a pig. Now, "If Ahmed was not a camel, Omar was not a pig." So Ahmed was turned into a camel. So, Abou was turned into a donkey.
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Nov 28, 2020 03:54
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Puzzle 55: Abou = donkey, Ahmed = camel, Omar = pig, Sharif = goat. Easily figured once you determine what Ahmed was turned into.
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Nov 28, 2020 04:07
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Neither Ahmed, Sharif nor Abou were pigs, so Omar must be the pig. This also rules out Ahmed not being a camel, so we have Ahmed the camel and Omar the pig. Abou wasn't a goat, so Sharif must have been the goat. This leaves Abou as the donkey.
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Nov 28, 2020 08:41
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Omar is the pig because everybody else isn't. Bc of that Ahmed isn't not a camel, so let's follow classic logic and make him a camel. Abou is a donkey because he isn't g,p or c. And then Sharif is the goat.
VictualSquid fucked around with this message at 12:58 on Nov 28, 2020 |
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Nov 28, 2020 12:54
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Correct, all! Puzzle 56: Beasts of Burden Three of the brothers, in their animal guises, were burdened with supplies for the town. They carried either kegs of oil or drums of dates. 1. If the donkey carried dates, then the goat carried oil. 2. If the donkey bore oil, then the camel carried dates. 3. If the goat carried dates, then the camel carried oil. Whose burden do we know? Who always carried the same load?
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Nov 28, 2020 21:02
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Puzzle 56: 1. If the goat carried dates, then the camel would be carrying oil and the donkey could therefore not be carrying either oil or dates. Thus the goat must be carrying oil. 2. After that, we only know that if the donkey had oil, then the camel carried dates. We don't know what the camel would be carrying if the donkey had dates. 3. Possible outcomes: C = dates, D = oil, G = oil; C = dates, D = dates, G = oil; C = oil, D = dates, G = oil
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Nov 28, 2020 21:50
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Whoops, missed 55. I didn't have any new and interesting answers to it anyway (other people covered the logic paths I would have mentioned), so on to 56. Puzzle 56: If the goat carried dates, he did not carry oil so clue 1 says the donkey carried oil. Then clue 2 says the camel carried dates, but clue 3 says the camel carried oil. That's an impossibility, and we can know that the goat carried oil. We cannot conclude anything about the donkey or the camel: the (donkey=oil & camel=oil) possibilities is forbidden by clue 2, but the other three possibilities are all allowed by the clues. So we only know the goat's burden.
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Nov 29, 2020 18:01
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Correct, both Ignatious and Tax! Puzzle 57: Feed Bags Abou, the donkey, had to share a stable and feed bags with a horse and a cow. 1. If About ate oats, then the horse ate what the cow ate. 2. If the horse ate oats, then About ate what the cow did not eat. 3. If the cow ate hay, then Abou ate what the horse ate. Who always ate from the same feed bag?
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Nov 30, 2020 00:50
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| # ? May 4, 2024 11:51 |
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Puzzle 57: Per 1, if Abou ate oats, the horse and cow ate either oats or hay, but both of those lead to contradictions. Therefore Abou ate hay. The horse and cow ate from the same feed bag as each other, but it's unknown whether they ate oats or hay.
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Nov 30, 2020 01:33
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