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#57 If Abou eats oats, then the cow and horse must both eat oats or both eat hay. If the former, then #2 is false. If the latter, then #3 is false. Therefore, Abou cannot eat oats. The question is unsolvable if there is an option other than oats or hay to eat, so Abou must always eat hay. The actual possibilities for Abou, horse, cow are : hoo, hhh, hho It's also unclear if it's meant to be 'who eats the same food always' or 'which animals eat from the same bag', except that only the first is answerable.
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Nov 30, 2020 07:08
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Correct, both of you! And now, we begin the final chapter of the tale of the genie and this merchant, with Magic Numbers. This one has a little bit of pre-puzzle plot! After ten years, the wives of Abou and his brothers appealed to the genie. "Sire, we beg you," said Sharif's wife. "Our husbands have suffered enough. And our children need their fathers." The genie agreed to transform the brothers back to their human forms, but only if the wives could give him three magic numbers which met certain conditions. Puzzle 58: The First Magic Number Here are the conditions of the first number: 1. If the first magic number was a multiple of 2, then it was a number from 50 through 59. 2. If it was not a multiple of 3, then it was a number from 60 through 69. 3. If the first magic number was not a multiple of 4, then it was a number from 70 through 79. What was the first magic number? Puzzle 59: The Second Magic Number Here are the conditions of the second number: 1. If the second magic number was a multiple of 6, then it was a number from 40 through 49. 2. If it was not a multiple of 7, then it was a number from 60 through 69. 3. If the second magic number was not a multiple of 8, then it was a number from 80 through 89. What was the second magic number? Puzzle 60: The Third Magic Number Here are the conditions of the third number: 1. If the third magic number was a multiple of 3, then it was a number from 50 through 59. 2. If it was not a multiple of 4, then it was a number from 60 through 69. 3. If the third magic number was not a multiple of 6, then it was a number from 70 through 79. What was the third magic number?
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Nov 30, 2020 07:26
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Puzzle 58: The First Magic Number LCM(3,4) = 12. There is no multiple of 12 between 50 and 59, so it cannot be a multiple of 2. It also cannot be a multiple of 4, since all multiples of 4 are multiples of 2. We now find a multiple of 3 between 70 and 79. 72, 75, 78, clearly only 75 is odd. 75 is our magic number. Puzzle 59: The Second Magic Number LCM(6,7,8) = 168. There are no multiples of 168 between 40 and 49, so it cannot be a multiple of 6. We now look for multiples of 7 (but not 6 or 8) between 80 and 89. 84 is the only multiple of 7 between 80 and 89, but is divisible by 6. We are now left with only multiples of 8. The only multiple of 8 between 60 and 69 is 64, which is divisible by neither 7 nor 6. 64 is our magic number. Puzzle 60: The Third Magic Number LCM(4,6) = 12. No multiples of 12 between 50 and 59, rule out a multiple of 3. Also rule out multiples of 6, since all multiples of 6 are multiples of 3. Find a multiple of 4 between 70 and 79 that is not a multiple of 3. 76 is our magic number. Aesculus fucked around with this message at 08:05 on Nov 30, 2020 |
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Nov 30, 2020 08:01
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Puzzle 58: The First Magic Number If the first magic number was a multiple of 2, then it would also have to be a multiple of 3 and 4 to avoid a contradiction in number ranges, but there is no such number between 50 to 59. It can't be a non-multiple of 3, because it would still have to be a multiple of 4 and therefore also a multiple of 2. Therefore it is a non-multiple of 4 and 2 that is a multiple of 3, and in the 70 to 79 number range, that means it must be 75. Puzzle 59: The Second Magic Number If the second magic number was divisible by 6, then it would have to also be divisible by 7 and 8, but this is impossible with the numbers 40 to 49. If it was indivisible by 8, it would have to be divisible by 7 while not being divisible by 6, and the only number from 80 to 89 divisible by 7 is also divisible by 6, so it is not indivisible by 8. The only remaining number from the only remaining range of 60 to 69 is 64. Puzzle 60: The Third Magic Number If the third magic number was divisible by 3, it would also have to be divisible by 4 and 6, and this is impossible with the numbers 50 to 59. If it was indivisible by 4, it would have to be indivisible by 3 but divisible by 6, which is impossible because 3 is a factor of 6. If it was indivisible by 6, it would have to be indivisible by 3 and divisible by 4, and of the numbers 70 to 79 the only remaining number it could be is 76.
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Nov 30, 2020 09:59
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Correct, both of you! And now, we have a section with a solid amount of plot! Hope these puzzles don't waterlog you, it's time to deal with The Dragon Montagne Puzzle 61: Seth Meets The Dragon Princess Fleur traveled to a distant mountain pool to sample its waters. There she and her retinue were captured by the terrible dragon Montagne, who had taken over the entire countryside. When word got back to the king, he offered half his wealth and the princess in marriage to anyone who could rescue her and defeat the dragon. One after another, the knights of the kingdom set forth. But each in turn failed. One day, Seth, a young peasant lad, went to face the dreaded monster. "My village is hungry and thirsty," Seth said. "We need the water you guard." "Seven times seven knights have failed to solve my seven puzzles and are now my prisoners," said the dragon Montagne, breathing out flames. "Would you be foolish, too?" "I'm only an ignorant peasant, not a knight," said Seth. "But I'll try." "Be prepared for the consequences!" thundered the dragon. "First, measure out exactly four cups of water in the pond using only these pitchers." The dragon swished his tail and two pitchers appeared. Seth picked them up. Neither measured four cups. One could hold exactly three cups and the other exactly five cups. Seth, however, worked on a farm. How did he measure out exactly four cups and thus complete the first task? Editor's note: This section sometimes has puzzles where you can only use the water provided in a pitcher, and other times, there is water available to fill at will. I will make sure to make note of this when it happens. For this, Seth has access to water, so he can fill or empty the pitchers as needed.
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Nov 30, 2020 23:55
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Puzzle 61 There are at least two ways to do it. One is: fill P5 from the pond, fill P3 from P5 (leaving 2 cups behind), dump out P3 and then put those 2 cups in it, refill P5 from the pond, and then fill P3 from P5, which uses 1 cup and leaves 4 cups in P5. Another one is: fill P3 from the pond, pour it all into P5, refill P3 from the pond, fill P5 from it (leaving 1 cup behind), dump out P5 and move that 1 cup into it, and then add another full P3 to it to get 4 cups.
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Dec 1, 2020 00:08
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Puzzle 61: Fill the 5 cup pitcher, then pour it into the 3 cup pitcher, then empty the 3 cup pitcher out and pour the last 2 cups in the 5 cup pitcher into the 3 cup pitcher. Fill the 5 cup pitcher again, and pour 1 cup into the 3 cup pitcher to fill it, and you'll have 4 cups of water in the 5 cup pitcher. e: There is another way but my way only wastes 3 cups of water, the second way wastes a lot more 
Ignatius M. Meen fucked around with this message at 00:16 on Dec 1, 2020 |
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Dec 1, 2020 00:13
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Correct~ Puzzle 62: The Second Challenge "That was just the first task," hissed the dragon angrily. With a swish of his tail, he dried up the mountain pond and made the two pitchers vanish. Suddenly, three new pitchers appeared, a five-cup, a seven-cup, and a twelve-cup pitcher. Then the dragon snorted and the twelve-cup pitcher was filled with water. The five-cup pitcher and the seven-cup pitcher remained empty. "Divide the water in the twelve-cup pitcher into two equal parts!" he challenged. How did Seth do it? Note: No emptying or filling. You have the twelve cups and that's it. To make up for this, trivia: "Fleur" is French for 'flower', and "Montagne" is French for "mountain".
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Dec 1, 2020 09:55
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Ooh, OK. Fill the 7-cup pitcher, then pour as much as you can into the 5-cup pitcher. (5, 2, 5) Empty the 5-cup pitcher back into the 12-cup pitcher (0, 2, 10), then pour the 2 cups into the 5-cup and refill the 7-cup pitcher (2, 7, 3). Next, pour as much as you can from the 7-cup pitcher to the 5-cup pitcher. (5, 4, 3). Pour the 5-cup pitcher back into the 12-cup pitcher and transfer the 4 cups from the 7-cup pitcher to the 5-cup pitcher. (4, 0, 8). Now, fill the 7-cup pitcher and pour 1 cup from it into the 5-cup pitcher. (5, 6, 1). Pour the 5-cup pitcher into the 12-cup pitcher (0, 6, 6). Done. Alternately, if you wanted to be cheeky, fill the 7-cup pitcher, then pour as much as you can into the 5-cup pitcher. (5, 2, 5) Then drink the two cups in the 7-cup pitcher. (5, 0, 5) Success! (assuming you don't get broiled)
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Dec 1, 2020 10:55
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Quackles posted:Alternately, if you wanted to be cheeky, fill the 7-cup pitcher, then pour as much as you can into the 5-cup pitcher. (5, 2, 5) Then drink the two cups in the 7-cup pitcher. (5, 0, 5) Success! (assuming you don't get broiled) No, this is still three parts. One of the parts is just in a different container, which is more human-shaped and oddly crispy.
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Dec 1, 2020 11:55
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#62- With the amounts after each pour indicated as (5-cup,7-cup,12-cup): 12->7 (0,7,5) 7->5 (5,2,5) 5->12(0,2,10) 7->5 (2,0,10) 12->7 (2,7,3) 7->5 (5,4,3) 5->12(0,4,8) 7->5(4,0,8) 12->7(4,7,1) 7->5(5,6,1) 5->12(0,6,6)
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Dec 1, 2020 22:00
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Correct, Kanga, Quack! Puzzle 63: The Three Steps The dragon Montagne breathed a bolt of fire. This time three jugs appeared. One of them, an eight-liter jug, was filled with water. The other two jugs, one measuring three liters and one measuring two liters, were empty. "Give me back four liters-in three steps." the dragon roared. And, in three steps, Seth gives him back four liters. How? Like the last one-no emptying or filling.
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Dec 2, 2020 10:04
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Trivial solution: Only the 8l jug can hold 4l. 4 = 8 - 4. 4 = 2*2. We simply pour from the 8l jug into the 2l jug (step 1), the 2l jug into the 3l jug (step 2) and then the 8l jug into the 2l jug again (step 3), and are left with 4, 2, and 2 litres.
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Dec 2, 2020 10:38
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Puzzle 62: Hopefully the shortening I'm applying will be reasonably intuitive... Fill 7 from 12, then fill 5 from 7. 5-5, 2-7, 5-12. Fill 12 from 5, then fill 5 from 7. 2-5, 0-7, 10-12. Fill 7 from 12, then fill 5 from 7. 5-5, 4-7, 3-12. Fill 12 from 5, then fill 5 from 7, then fill 7 from 12. 4-5, 7-7, 1-12. Fill 5 from 7, then fill 12 from 5. You now have 6 cups of water between the 7 and 12-cup pitchers. Puzzle 63: I was expecting that to be harder... kind of disappointed after digging into it. Fill 2 from 8. 6-8, 0-3, 2-2. Fill 3 from 2. 6-8, 2-3, 0-2. Fill 2 from 8. 4-8, 2-3, 2-2. Hand over the 8-cup pitcher and that's it. e: Nidoking posted:No, this is still three parts. One of the parts is just in a different container, which is more human-shaped and oddly crispy. But if you burn me up, the water will evaporate, and there will only be two equal portions left of liquid water in the other containers. And if the water vapor and air count as a third portion + 'container', then it's already impossible to beat your task anyway. 
Ignatius M. Meen fucked around with this message at 11:07 on Dec 2, 2020 |
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Dec 2, 2020 10:59
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Ignatius M. Meen posted:But if you burn me up, the water will evaporate, and there will only be two equal portions left of liquid water in the other containers. And if the water vapor and air count as a third portion + 'container', then it's already impossible to beat your task anyway. At that point, the containers from all of the previous challenges have been destroyed, so you already fail them all retroactively. It doesn't work that way. And yes, I will deconstruct every joke into its component atoms and scatter them to the winds, see if I don't.
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Dec 2, 2020 11:41
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Correct! Puzzle 64: Wicked Walter What no one knew was that the dragon Montagne was really the wicked apprentice wizard Walter in disguise. He had great natural talent as a wizard, fooling the entire kingdom with his dragon act, but so far the only things Walter had learned to conjure up were containers and the only puzzles he knew had to do with measuring water. He didn't have power over anyone who could solve his puzzles, and though he feared Seth would continue to succeed where others had failed, he desperately continued with his bluff. Walter (as the dragon) conjured up three urns. One contained ten gallons of water. The other two were empty, one capable of holding four gallons, and the other three gallons. "Give me back five gallons, using only 5 steps." Walter demanded of Seth. Just like before-no emptying or filling.
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Dec 3, 2020 08:52
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Puzzle 64: Wicked Walter from 10, pour 4 gallons into 4; from 4, pour 3 gallons into 3; from 3, pour 3 gallons into 10; from 4, pour 1 gallon into 3; from 10, pour 4 gallons into 4. Also, since the dragon can't barbecue anyone, do the second solution I had to that other puzzle for shiggles.
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Dec 3, 2020 10:08
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Puzzle 64: 1. Fill 4 from 10. 2. Fill 3 from 4. 3. Fill 10 from 3. 4. Fill 3 from 4. 5. Fill 4 from 10. Hand over the 10. As for Walter, his fire spells probably don't use conjuration, so that and the general lack of education in the kingdom is probably how he's been selling his dragon act.
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Dec 3, 2020 15:01
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Correct! Puzzle 65: Drop by Drop Snarling, Wicked Walter (as the dragon) materialized two vials so small that Seth could place them both on one finger. One vial could hold five drops of liquid and the other seven drops. "What is the least number of steps it will take to give me three drops of water, and what is the least number of steps it will take to give me four drops of water?" challenged the dragon. "Aren't you giving me two puzzles?" protested Seth. Walter snorted fiercely. Seth set about solving the problems. How? Like the first problem-you can fill or empty at will with this one.
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Dec 4, 2020 01:10
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#65a 4 steps. Fill 5, pour into 7, repeat. There will be three drops remaining in the 5-drop vial. #65b 6 steps. Fill 7, pour into 5. Pour out the 5 and move the two drops from the 7 into the 5. Fill the 7, pour into 5 and there are 4 in the 7-drop vial. And then fail due to surface tension making it impossible to do any of that.
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Dec 4, 2020 04:53
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Puzzle 65a: Fill 5, fill 7 from 5, fill 5, fill 7 from 5. 3-5, 7-7, 4 steps. Puzzle 65b: Fill 7, fill 5 from 7, empty 5, fill 5 from 7, fill 7, fill 5 from 7. 5-5, 4-7, 6 steps.
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Dec 4, 2020 10:26
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Correct, for both! Puzzle 66: Triple Threat For the next challenge, Wicked Walter placed before Seth four jars, the largest of which held nine liters of water. The three empty jars could hold five, four, and two liters respectively. "This time," Walter said: "You are to divide nine liters into three equal parts." How many steps does it take Seth this time?
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Dec 5, 2020 00:25
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I've got a method that does it in six steps, but I'm sure there's a shorter way. 1-2) Fill the four and two liter jars from the nine liter jar, leaving three liters in the nine liter jar. We won't be touching that one again. 3) Pour the two liter jar into the five liter jar. 4) Pour the four liter jar into the five liter jar, leaving one liter in the four liter jar. 5) Pour the five liter jar into the two liter jar, leaving three liters in the five liter jar. 6) Pour the two liter jar into the four liter jar, making three liters in each jar that can hold that much.
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Dec 5, 2020 00:51
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Puzzle 66: Fill the 5 from the 9, 4-5-0-0 Fill the 2 from the 5, 4-3-0-2 Pour the 5 into the 4, 4-0-3-2 Pour the 2 into the 5, 4-2-3-0 Fill the 5 from the 9, 1-5-3-0 Fill the 2 from the 5, 1-3-3-2 Move the 2 back into the 9, 3-3-3-0 Seven steps. I see Nidoking found a six-step method, which is more or less the same as mine except after my step 4 I've reached his step 3 (his step 3 ends up with 3-2-4-0 and mine ends up with 4-2-3-0), so his is better. I suspect there isn't a five-step method, but I can't prove it yet.
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Dec 5, 2020 05:21
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Puzzle 66: 1. Fill 2 from 9, 2. fill 4 from 9, 3. fill 5 from 4, 4. fill 5 from 2, 5. fill 4 from 2, 6. fill 2 from 5, 7. fill 4 from 2. 3-9, 3-5, 3-4, 0-2, 7 steps. Hm, guess I have to sit in the wrong box with Tax Refund. Should have thought through the step after getting 3 liters in the 9 liter jar more.
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Dec 5, 2020 07:19
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Ooh, boy, this'll be fun. If any water you drink counts as an additional part: Fill the 5-liter jar from the 9-liter jar. (0/0/5/4) Fill the 2-liter jar from the 5-liter jar. (2/0/3/4) Drink the 3 liters of water that are in the 5-liter jar. (There's no time limit, so take your time.) (2/0/0/4 ; 3) Pour the 2-liter jar into the 5-liter jar. (0/0/2/4 ; 3) Fill the 5-liter jar from the 9-liter jar. (0/0/5/1 ; 3) Fill the 2-liter jar from the 5-liter jar. (2/0/3/1 ; 3) Pour the 2-liter jar into the 9-liter jar. (0/0/3/3 ; 3) You have 3 equal parts of 3 liters. If it doesn't: Fill the 5-liter jar from the 9-liter jar. (0/0/5/4) Fill the 2-liter jar from the 5-liter jar. (2/0/3/4) Drink the 3 liters of water that are in the 5-liter jar. (There's no time limit, so take your time.) (2/0/0/4) Pour the 2-liter jar into the 5-liter jar. (0/0/2/4) Fill the 2-liter jar from the 9-liter jar. (2/0/2/2) You have 3 equal parts of 2 liters.
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Dec 5, 2020 09:04
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Nidoking got it right, the lowest is indeed six steps! And now, the final water riddle... and it's a doozy. Puzzle 67: The Rescue For the final challenge, Walter, the dragon Montagne, presented Seth with two ten-gallon vats full of water and two pitchers, one holding five pints and the other holding four pints. "Place one quart in each pitcher," Walter said. After several hours, Seth wearily presented the dragon with a quart in each pitcher, whereupon the dragon Montagne was revealed to all as the Wicked Walter. He had no choice but to free Princess Fleur, her ladies-in-waiting, her pages, and the forty-nine knights who had taught to rescue her. How does Seth solve this last puzzle and expose Wicked Walter? You have pre-provided water for this one but you can empty as you see fit. Also, just to make sure you don't have to look it up: 1 Gallon is equal to 4 Quarts, and 1 Quart is equal to 2 Pints-correspondingly, 1 Gallon is 8 Pints.
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Dec 6, 2020 04:56
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Think I've got this one too, and if I'm right, it IS a doozy. Fill the five-pint pitcher and pour it into the four-pint pitcher to isolate one pint. Empty the four-pint pitcher back into the vat. Now... just turn one of the ten gallon vats and dump the whole thing. Just all over the ground. So you can pour that one pint into it. Now just repeat the first three steps with the other ten gallon vat. You can then pour that one pint from the nearly empty vat back into the four-pint pitcher. Whoops, can't measure properly or read those vital units. I'll leave it up because I'm amused, but it won't solve the puzzle. Okay, how about this: Same idea as above, get one pint into one of the vats, then repeat with water from the other vat and add that, so you've got two pints (one quart) in a vat and tons of water left in the other. Fill the five and pour it into the four once again, dump the four back into the vat, and put the one remaining pint from the five into the four. Now you can refill the five, dump that into the four, leaving two in the five. It's just a matter of emptying the four and transferring the two from the other vat back into it. More steps, but the same gimmick of emptying one of the vats to hold one of the target amounts. Nidoking fucked around with this message at 05:16 on Dec 6, 2020 |
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Dec 6, 2020 05:10
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Puzzle 67: Empty 10-2. Empty 4, fill 4 from 5, empty 4, fill 4 from 5, fill 5 from 10-1, fill 4 from 5. Fill 10-2 from 5, repeat steps 2-7, empty 4, and fill 4 from 10-2. You now have 2 pints in both the 4 and 5 pint pitchers.
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Dec 6, 2020 06:24
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Puzzle 67: Using one of the vats, fill the 5-pint container 10 times and dump it out each time. Do this 4 times with the 4-pint container and the same vat. There are now 14 pints remaining in the vat. Fill the 4-pint, pour it into the 5-pint. Do this again, and when the 5-pint is filled, there will be 3 pints in the 4-container and 6 in the vat. Fill the 4-pint container once more, and pour 2 pints to fill the 5-pint, leaving 2 in the 4-pint container. Dump out the 5, and fill it with the last 2 from the vat. So why two vats?
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Dec 6, 2020 08:36
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Puzzle 67: Two 10 gallon (80 pints) vats: VA, VB. Two pitchers: P4 (4 pints), P5 (5 pints). Goal 1 quart (2 pints) in each pitcher. Values put in pints. Step 1: Dump VB. gently caress it. You need the room. VB = 0 Step 2: Fill P5 from VA. VA = 75, P5 = 5 Step 3: Fill P4 from P5. P5 = 1, P4 = 4 Step 4: Dump P4. P4 = 0 Step 5: Fill VB from P5. VB =1, P5 =0 Step 6: Repeat Steps 2-5: VA = 70, VB = 2 Step 7: Fill P4 from VA and dump it 17 times. 17*4 = 68, 70-68 = 2. VA = 2 Step 8: Fill P4 and P5 from VA and VB respectively. P4 = P5 = 2
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Dec 7, 2020 07:50
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Huh. Nobody was really 'wrong' but none matched the way it was done in the book-I'll provide it for posterity. But anyway, we might have the most non-logical and pure math-y chapter of this book-The Wizards of Odds. I'll save time and just give all the riddles in this chapter in one batch. Some of them do come in sets so you can quickly answer all of them. 'Puzzle' 68: The Well of Wisdom Once a year, apprentice wizards, witches, sorcerers, and sorceresses come from many different kingdoms to a conference at which they learn about new potions and omens. This year they meet in the kingdom where Merlin serves as Chief Wizard. Of the thirty apprentices, he invites, only two are unable to attend, having been wounded fighting off a sudden influx of dragons. As the apprentices come in, Merlin gives them each a gold coin to toss into the Well of Wisdom. If the toss of the coin matches the previous toss--whether it was heads or tails--the apprentice who pitches it will acquire a new power. Wizard Merlin will toss the first gold coin. What chance is there that it will come up heads? 'Puzzle' 69 (nice): Evelynne at the Well The first guest to arrive is Evelynne, apprenticed to the lady of the Lake. What is the probability that Merlin's and Evelynne's tosses both come up heads? 'Puzzle' 70: And Percival Makes Three Percival, aide to the Wizard of the Woods, tosses the next coin. What is the probability that all three coins will come up heads? 'Puzzle' 71: Four Coins in the Well Vivienne, student of the Woodland Sorcerer, tosses fourth. What are the odds for heads coming up in all four tosses? 'Puzzle' 72: Oberon's Toss Supposed Vivienne's coin comes up tails. What are the chances that Oberon's coin, tossed right after Vivienne's, comes up tails? 'Puzzle' 73: How Many Wiser Wizards? How many of the twenty-eight apprentices are likely to match the previous toss and acquire the new power Merlin promised? 'Puzzle' 74: Magic Seeds Merlin passes around an urn of new black and white magic seeds to the budding wizards and witches for them to sample. The magician who developed them claims that the black seed makes one impervious to dragon fire and that the white seed provides the cloak of invisibility. The urn dispenses one seed to each apprentice sorcerer. By the time it reaches Titania and Garth of Glend there are only two black seeds and one white seed left. What is the probability of both getting a black seed? 'Puzzle' 75: Blind Sorcerer's Buff Merlin singles out four of the apprentice sorcerers and asks each one to stand at a corner of the huge ballroom in the castle. He blindfolds them and steers them to the middle of the room. Then he turns each around several times and tells them to return to their original corners. IF each one winds up in a corner, what is the probability that all four will succeed in getting back to the right one? 'Puzzle' 76: Encore To vary the challenge, Merlin blindfolds another four apprentices and sets them in the four corners of the ballroom. Then he steers them, blindfolded, to the middle of the room, directing them to return to their original corners one at a time. Once a corner is occupied it is off-limits. If each one reaches a corner, what are the odds that these four apprentices will return to their original corners? 'Puzzle' 77: Non-Magic Magic Merlin calls on two of the apprentice sorcerers to entertain with magic tricks using an ordinary deck of cards. Unfortunately, neither apprentice has practiced enough to be very good at card tricks. They are at the mercy of the laws of chance. If each one pulls a card at random from a deck of fifty-two cards, who has the better chance of being a successful magician: Lorelai, who promises to come up with one of the four aces? Or Urth, who claims his first card will be one of the thirteen hearts? 'Puzzle' 78: More Card Tricks Will Lorelei have a better chance of coming up with two aces if she returns the first ace she draws to the deck or if she puts it aside? 'Puzzle' 79: Hearts for Urth Will Urth have a better chance of coming up with two hearts if he returns the first heart to the deck or if he puts it aside? 'Puzzle' 80: Go Fish Merlin provides some fun by stocking the castle pond with a fish for each guest. Two of the thirty fish he endows with enchantment. The student wizards and sorcerers take turns casting. Does Pendragon, apprentices to the not-so-wily wizard of Trelawn, have a better chance of luring an enchanted fish if he goes first or if he casts tenth and one enchanted fish has already been caught? 'Puzzle' 81. Two Enchanted Fish? What are the chances that Pendragon and Elaine of Camelot both get enchanted fish? Yeah this section isn't puzzles or riddles, it's the probability section of a math text book dressed up in BBC's Merlin.
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Dec 7, 2020 08:20
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Can't seem to load this. 'Puzzle' 68:  1 in 2 or 50%, of course.'Puzzle' 69 (nice): 1 out of 4 or 25% per [HH, HT, TH, TT]. 'Puzzle' 70:  1 in 8.'Puzzle' 71: 1 in 16... 'Puzzle' 72: Ah, we're not mentioning the other coins this time. 1 in 4. Congrats, I had to wake up to answer that one. 'Puzzle' 73: Talk about whiplash. I'm probably not looking hard enough for the right formulas but I'm not interested in doing the author's actual math homework problem either. 'Puzzle' 74: This is more reasonable, if very easy. 1/3 out of [BB, BW, WB]. 'Puzzle' 75: 1/4 * 1/4 * 1/4 = 1/64. 'Puzzle' 76: 1/4 * 1/3 * 1/2 = 1/24. Misread 75 and thought this was the answer to it, then was a bit confused when I read this one. 'Puzzle' 77: Urth because he's got the 1/4 chance vs. Lorelai's 1/13 chance. You don't even have to do any math for this answer, this entry tells you there are more hearts than aces in a deck! 'Puzzle' 78: ....I can't tell if this is supposed to be 'clever' but it just comes off as plain stupid. Obviously she'll have a better chance of drawing/'coming up with' a second ace if she returns the first one to the deck, it just won't be a different ace, and the wording isn't requiring that. 'Puzzle' 79: see above, replace she with he and ace with heart 'Puzzle' 80: 2 out of 30 or 1/15 chance vs. 1/21, he should definitely go first with these choices. 'Puzzle' 81: Welp, I knew it would happen eventually with this nonsense level. There's no way to be sure how to answer this question because we have no idea when the hell Elaine gets her turn in the lineup (presuming Pendragon will follow the advice of #80 is a leap, but at least it's there). Assuming that Pendragon is going first per our rec in entry #80, and that Elaine is going second, that's 1/15 * 1/29 = 1/435. Ignatius M. Meen fucked around with this message at 10:41 on Dec 7, 2020 |
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Dec 7, 2020 10:39
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I agree with most of the above, but... Puzzle 72: The probability is 1/2. We're given that the previous flip was tails, so that doesn't affect the probability in the least. The probability of the next flip being tails is 1/2. Puzzle 73: There are 28 flips, each with a 1/2 chance of matching the previous flip. The expected number of matches is half of 28, or 14. Each flip is technically independent, since it doesn't matter whether the previous flip was a match or not. It's only which face was flipped that determines the target face for the next flip. Puzzle 75: I also wondered whether the apprentices could return to the same corner until I read the next puzzle. But since all four apprentices could well end up in the same corner, I have (1/4) ^ 4 or 1/256 as the probability of all four ending up in their starting corners. Puzzle 81: This type of sequential probability actually doesn't depend on the order at all, since there are as many fish as there are fishers. I have no idea who the thirtieth fisher is, since it's Merlin and 28 apprentices, but it says "one fish per guest" and I take that at its word. Since each guest gets a fish, two will get the enchanted fish, and it's just a matter of which two those are. That's one divided by (thirty choose two), which is 30! / (28! * 2!) or one divided by (15 * 29), which is the answer you gave. Taking it in sequence as you have, you'd find that by fishing later, there is a probability that one or both enchanted fish will have been caught before their turns, zeroing out the probability in that case, but that adds to the increased probability they have of catching the enchanted fish if they haven't already been caught, and it will balance at 1 / (15 * 29) every time. Probability is pretty cool that way.
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Dec 7, 2020 14:17
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I was curious whether the vat problem was meant to imply you couldn't drain the vats, but it didn't say anything about it. In 81, there are 30 fish because there were 30 originally expected, but only 28 showed up. That complicates things slightly. You can still use C(30,2), considering the 'pond' as taking two turns at getting a fish. So it's still 1/435 The reasoning below is wrong because the cases where the pond has one fish left are more likely. In essence, they should just be left in and the numbers still work out. e: Actually, it doesn't change things, which was my first thought, but I was half-awake and second-guessed myself. Kangra fucked around with this message at 19:13 on Dec 7, 2020 |
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Dec 7, 2020 17:38
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Thanks to the tag-team of Ignatius and Nidoking, all are gotten right (in case you're curious, #81's correct answer was 1/435. And now, I am proud to announce that we have reached the second half of the book-Logic From Far Away. So we're done with fantasy puzzles and now we begin with science fiction puzzles! We begin our trek through the stars on The Planet Dranac- The Planet Dranac lies far beyond the known solar system. It has its own unique beings and civilization. It is apparent that Dranac is far behind Earth in technological development. As well, Dranacians have less respect for honesty. Deception and crimes are commonplace among the inhabitants. Puzzle 82: Duplgoose Eggs Duplgooses, small farm animals, lay eggs only occasionally, although in pairs. Their eggs, an important food source on Dranac, are a highly valued commodity. A basket of duplgoose eggs has been stolen from the open market in the local village. There are four suspects, and each makes one statement, although only one of the four speaks truthfully. The guilty one can be deduced from their statements below. A said, "B did it." B said, "D did it." C said, "I did it." D said, "Either A or C is the guilty one." Did C do it? If not, who did?
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Dec 8, 2020 09:41
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If A is telling the truth, then D did not do it, C did not do it, and neither A not C did it. B did it. If B is telling the truth, B did not do it, C did not do it, and neither A nor C did it. D did it. If D is telling the truth, then B did not do it, D did not do it, and C did not do it. A is the only remaining person who did it. In none of these scenarios did C do it.
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Dec 8, 2020 09:48
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If C is telling the truth, then D is telling the truth also. Which means he lies. If A is telling the truth, B did it. If B is telling the truth, D did it. If D is telling the truth, A did it unless C is also telling the truth. So C is guaranteed innocent. No Idea on how to narrow it down further.
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Dec 8, 2020 14:27
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Puzzle 82: If C did it, then C and D would be telling the truth, and thus contradict the setup, so C is a liar and did not do it. D accuses A (and C, but C is exonerated), A accuses B, and B accuses D, which means that any of them could be guilty and only one of them would be telling the truth. So no and A, B, or D respectively.
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Dec 8, 2020 14:45
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| # ? May 4, 2024 06:43 |
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Well, you all are partially correct, according to the book: it says that A did it, and D is telling the truth. I'll give you that one. Puzzle 83: Huffalon Thefts Huffalons-great sturdy beasts that are used for riding, pulling carts, and carrying bundles-are the chief source of transportation on the planet. There have been a number of huffalon thefts in the local village. It is apparent that there are two thieves working together. Several suspects have been identified, and two of them are the guilty ones. Each makes one statement below, and four of the six statements are truthful. Only the two guilty parties make false statements. A said, "C is guilty". B said, "F is not guilty." C said, "D is not guilty". D said, "B is guilty." E said, "A is not guilty." F said, "E is not guilty." Which two are guilty?
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Dec 8, 2020 23:17
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