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I read the previous LP but don't remember the puzzles, so I'll give this a shot. (It's possible that I might remember future puzzles, in which case I'll recuse myself from those.) Puzzle 002: The clues are annoyingly vague; it's hard to tell what counts as "nearby" for the sound of music. Does two windows over count? If so, one good candidate is the most upper-left window, that has a blue flag in front of it. I also wonder whether the flag being "in front of the window" means it has to be hung from the same surface as the window, or whether the flag needs to be visible by looking out the window. My money's on the latter, so I think the answer is the lit window just above the one with the cat, that faces the right-hand side of the picture. It is adjacent to the music, and the red flag hung below the music window is visible from that one. It's also one of only two windows where a flag is visible at all. Puzzle 004: I'd circle the missing curtain on the left side of the window in the third picture. It could easily have been torn off by whoever was in here, who then tied it to something (most likely the tail of the Kronosaurus) and went out the open window. The curtain's not long, but it's long enough to reach one window below. If that window was open (as it is in the previous picture if my earlier guess was right, since it would be the one with the cat), our mysterious burglar could just go in that apartment and let himself out by its front door. Assuming that it was open, of course. EDIT to add: The other possibility is the diagonal bookcase in picture 4; someone could be hiding behind there right now for all we can tell.
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Dec 27, 2015 05:20
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Puzzle 017: Rather than a textual description, here's a diagram of what I believe is the longest possible route: http://lpix.org/2322819/037-solved.png Puzzle 024: In each case, if you had folded it 0.5 cm further left or right, the fold would have been perfect. And logically, that 0.5 cm line would be halfway between the two fold lines. Therefore, the fold lines are 1 (one) cm apart. Puzzle 023: If you put the jug at A, every odd-numbered move will place the vinegar in jug 1. If you put the jug at B, the odd-numbered moves will have the vinegar in jug 2. Since there are 55 moves and you want 2 to end up with the vinegar, put the pitcher down in spot B. Puzzle 007: The only one with a mustache is B, so the O'Connors are A and B. The Joneses are sitting by the aisle, so they're across from each other in spots C and G. The Lambers are also opposite each other, so they're in D and H, so the Hadleys must be in the two remaining spots, E and F. Mrs. Hadley is in spot E.
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Dec 30, 2015 15:20
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The hamster is so fat, he's a Porkster! Puzzle solutions: Puzzle 050: http://lpix.org/2337826/050-solved.png Puzzle 010: The area in the lower left has three areas that all touch each other, so you need a minimum of 3 colors. The rest of the canvas could be colored with just two colors, since most of the intersections are four-way corners which don't spoil a "checkerboard-like" coloration. So although the theoretical maximum is 4 colors for any 2-D painting or map, you need a maximum of 3 colors for this particular map.
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Jan 17, 2016 19:05
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Glazius posted:I do question that posh lady's dress sense. Could you spoiler-text your answers? I wanted to solve the puzzles, but by the time I'd scrolled down to click on Reply, I saw your answers and now I can't solve the puzzles unspoiled.
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Feb 1, 2016 00:40
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025: I remember this one from last time, but I'll solve it again for fun. Only two animals have a path to the exit: the lion and the rabbit. However, the question asks which animals will make it "safely out of the zoo", and only the lion qualifies. The rabbit will technically make it out of the zoo, but inside the lion's belly, which doesn't count as "safely". 021: Three minutes. Minute 1: Person A tells the message to person B. Now two people know the message. Minute 2: A tells it to C, B tells it to D. Mine 3: A through D tell it to E through H respectively, and now all eight people know the message.
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Feb 15, 2016 20:01
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The different versions of puzzle 52 are the same puzzle, just with a different label on "day 1" of the week. So I'll present just one solution, calling the days "day 1" through "day 7". (And that's why they split it between European and US versions, I guess: so that Americans wouldn't say, "Hey, wait a minute, Monday through Sunday is six days of one week and the first day of a second week" or Europeans wouldn't say "Hey, wait a minute, Sunday through Saturday is the last day of one week and six days of a second week.") 052: It was cloudy on day 3, and he didn't wear his hat. If it had been cloudy on day 2, he would have worn his hat on day 3 as it would have been the second cloudy day in a row. Also, if it was cloudy on day 4, he would have worn his hat then, and he didn't. So day 2 and day 4 are solved: hat on day 2 means it was sunny, no hat on day 4 means it was rainy. We have one more rainy day to place, and day 6 is the only remaining non-hat day, so day 6 is where that last rainy day goes. Now we have: _, S, C, R, _, R, _ from day 1 to 7 (_ means "don't know yet"). On all three unknown days, he wore his hat -- which means days 5 and 7 can't have been cloudy, because then he'd have been breaking his rule. So day 1 was cloudy, and it was the second cloudy day in a row (day 7 of the previous week must also have been cloudy). Therefore, from day 1 to day 7, the weather was: cloudy, sunny, cloudy, rainy, sunny, rainy, sunny. 051: It can't be C because the curly whisker of the top cat should align with the edge of the vertical stripe behind it. It can't be A because the rightmost eyelash of the top cat should be long. So B is the correct inverted reflection. 038: Move 1 and 2 to point B (2 hours so far). Ride 1 back to A (3 hours so far). Move 4 and 6 to B (9 hours so far). Ride 2 back to A (11 hours so far). Move 1 and 2 to B, done in 13 hours.
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Mar 19, 2016 10:12
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033: You could do a thing with lines drawn from outside the net, where if you cross an odd number of rope lines to get to a fish it's inside, and if you cross an even number of rope lines it's outside the net. But it's easier to notice that the fish directly in front of where the ropes enter the water is definitely in the net, and then just circle all the fish you can reach from that one without crossing any lines. All the other fish you could reach by crossing one line, so they're outside. Result: http://lpix.org/2405755/033-solved.png, with 7 fish inside the net. 044: Punting on this one. I've never been any good at visualizing knots. 042: That turned out easier than I thought it would be: http://lpix.org/2405757/042-solved.png and slide the top piece one square to the right and one square downwards. 043: This one was a little trickier, but once you start looking at the shape of the holes, the solution becomes a bit easier to see: http://lpix.org/2405758/043-solved.png and then take the "upper" piece and, without rotating it, slot it into the hole of the "lower" piece. 050: The double negatives are confusing, so let's replace each statement with one that's exactly equivalent: A: D is lying. B: C is lying. C: A is truthful. D: E is lying. E: B is lying. We know from the puzzle introduction that there are two truthful cows and three liars. C is the only one who said another cow was truthful; if C is telling the truth, then so is A. That means D is lying, so E is truthful -- but now we have three truthful cows, and we know that's not the case. So C is lying, and therefore so is A. So D is truthful, E is lying, and B is truthful. The lying cows are A, C and E. 048: Only one person has a beard, hat, cane AND bowtie: the man on the upper left, with the blue beard, pinkish hat and white gloves. Edit: I just looked at the hint coin for this one. I got it right, but that isn't a pinkish hat he's wearing. It's his bald head, because his hat (the brown one in the upper left corner of the picture) was just blown off his head by the strong wind, despite his attempts to hang on to it.
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Mar 27, 2016 18:52
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^^^^^ Algebra's wrong on 031. Second equation should be 3(T-2) = J+2. ^^^^^ 039: C ended up with red trousers from A, so A must have kept his red shirt. And since nobody's shirt or trousers matched, C ended up with a blue shirt from B, so B must have kept his blue pants. Therefore, B ended up with C's white shirt, and A ended up with C's white pants. The new outfits are: A - Red shirt, white pants B - White shirt, blue pants C - Blue shirt, red pants 031: I'll skip the algebra. Johnny has 7, Thomas has 5. 046: C fits on the bottom, then A and B easily form the top layer. D is the odd shape out. 035: D is not red (clue 4). It can't be blue either (clue 1). It can't be green (clue 3). So D is yellow. The only one farther from the lake than D is A, so by clue 2, A is green. That leaves red and blue in B and C. C is closer to the pine tree than B, so C is red and B is blue. Results: A is green, B is blue, C is red, D is yellow. 054: Possible color patterns with no repeated color: ABC, ACB, BAC, BCA, CAB, CBA. Patterns with one repeated color that are allowed: ABA, ACA, BAB, BCB, CAC, CBC. And that's it. 12 designs. 040: Found time to solve this now. My answer: http://lpix.org/2414196/040-solved.png 049: The telephone is useless unless more people have one. 055: Take one of those chains and open every one of its seven links. There are now seven chains remaining, and seven "gaps" to be filled; fill each gap with one of those seven links. Total cost: £14. Tax Refund fucked around with this message at 11:00 on Apr 3, 2016 |
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Apr 3, 2016 06:00
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Glazius posted:Everybody's been pretty good so far, but I think the flag's a bit different from what people are saying. You know, I think you're right. If white isn't allowed as a color, then following your reasoning, the answer works out to 3 * 2 * 2 = 12, which is what I came up with. But you're right; white is totally a legitimate flag color. Therefore, I'm changing my answer to 36.
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Apr 4, 2016 03:59
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Dragonatrix posted:I am unbelievably envious of how easy people that aren't me seem to have found this one. Teach me your ways, magical people. Actually, that one was NOT easy at all for me. It might look easy because you're just seeing the final answer instead of the hour I spent waving my hands in the air tracking folds, and which edge ends up where. Now for solutions to this set of puzzles: 056: The X is in position 2. Moon is 2 below diamond, so moon can't be in 4. Moon and diamond are in 5 and 3, or 3 and 1, respectively. Since the circle must be above the diamond, diamond can't be 1, so moon and diamond are in 5 and 3. Star is next to moon, so star is 4 and circle is 1. From top to bottom, then: circle, X, diamond, star, moon. 057: Red dots represent rose positions: http://lpix.org/2421467/057-solved.png W06: Inscribe a circle inside the square, then inscribe a square inside the circle. Result will be a diamond whose points are the midpoints of the original square's sides, which is half as big as the original. The answer is 2, since the original square was twice as big as the final square. W07: A must be an even number, since B + B = 1A. B must be 1 greater than A, so B is an odd number. And B must be at least 5 or it wouldn't have carried. 5 + 5 = 10, and A = 0 doesn't work. 7 + 7 = 14, and A = 4 doesn't work. 9 + 9 = 18, and that works. B = 9, A = 8, and the equation was 89 + 9 = 98. W08: For each pair of adjacent numbers in one row, add together all their digits to get the number below that pair. So for 160 and 144, 1+6+0 + 1+4+4 = 7 + 9 = 16, and so on. The ? represents 1+6 + 1+8 = 7 + 9 = 16, as well.
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Apr 8, 2016 12:46
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059: This one actually stumped me during the first thread. And since I remember its solution, I'm going to recuse myself from solving it this time. The others, though, I don't remember, so I'll solve them. W10: Can't be A; the lower left chicken has its eye closed in A, but open in B and C (and in the flipped original). And can't be C; the lowest chick is facing left in C, but right in B and C (and left in the flipped original). So B is the right answer. W11: I'll skip the map I drew. Territory C will be blue. W12: C3 moves to B2 (the center). Now white threatens a horizontal line on row 2, and a diagonal line on A3-B2-C1. Black can only block one of those two threats, and white will win next turn. W13: I remember from my childhood math books that the answer is C, but I still have no idea how that works. W14: This one is a classic. Unless you're starting or ending on an island, you'll have to enter AND leave it by different bridges, so it would have to have an even number of bridges. If any island has an odd number of bridges, there's no way to traverse all of its bridges once unless you either started on, or ended on, that island. If there are just two odd-numbered islands, then make one the start point and the other the end point and you'll win; if there are more than two such islands, you can't win. Since there are only two odd-numbered islands in this map, and the goal is one of them, it's possible -- but only if you start at the other odd-numbered island, which is A. Note that it's not even necessary to trace the bridge routes; just look for the odd-numbered islands.
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Apr 14, 2016 08:33
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Max Peck posted:W13: B! In addition to only having one 'side', a Mobius strip only has one edge--trace along it and see. Since we never cut that edge when we cut the Mobius strip down the middle, it must still be one continuous edge in the resulting shape. In other words, we can't wind up with two separate loops, despite having cut it in half. B is the only shape like that. You're right. C is the result of cutting B in half, but B is the result of cutting the original Möbius strip in half. Changing my answer for W13.
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Apr 15, 2016 00:43
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062: http://lpix.org/2443063/062-solved.png W15: You're essentially flipping the square one 90° turn to the left each time, which is disguised by the sides of the polygon being angled. Since you'll make five left rotations, that's the equivalent of one left rotation, and you'll end up with the arrow pointing left, as in B. W16: I'm having a very hard time visualizing this one. I'll solve it later.
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May 1, 2016 01:44
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089: Divide the space between the beds into four by drawing two diagonal lines through it. Now each of those four sections forms a "cap" that you can stick onto a quarter of the circle at the bottom, and you end up with a diamond shape (a square rotated 45 degrees). The circle's radius forms a line between the center of the square and the center of one side, so each side of that square is 20 meters. Therefore, the area colored red is 400 square meters. 068: I remember this one from before, so I'm going to recuse myself from answering it. 072: A is probably last, since the neon light in the sign burned out. B is next-to-last since the window is broken, and it was whole in C and D. (In A, it was covered over with paper as a temporary repair measure, so A came after B). In C the lamp is off and in D the lamp is on; it remains on through A and B, so it's likely that C came first. Then the lamp was turned on in the evening, producing D, B and A. So in order: C, D, B, A.
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May 9, 2016 11:42
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117: The sister's husbands are C (no hat - married to the younger sister) and A (no mustache - married to the elder sister). B is therefore the one married to neither sister, and you don't actually need to figure out who the sisters are at all! 070: The stones' values are 100 through 1200 in steps of 100. So their total value is (100+1200)*6 = 7800, and we need to find three chains worth 2600 each. At the top, I see 900 next to 700, and there's a 1000 next to it, so that's one chain. 1200, 100, 500, 800 make the second chain, and 1100, 200, 400, 300 and 600 make the third chain. Cut between 900 and 1200, between 800 and 1100, and between 600 and 1000. 093: My sister is twice my age, and in five years, I'll be her age. So I'm 5 and she's 10. (And we didn't need to work out Mum and Dad's ages at all, but she's 30 and he's 35.) 063: Total sum of all cards is 21, so the top row needs to add up to 10 and the bottom row to 11 (clue 1). The 5 must be to the left of the 4 (clue 3); if the 4 is in the right column, then the 1 must be in the right column as well to make it add up to five (clue 2). But 1 must be in the left column (clue 4), so the 4 can't be in the right column. So the 4 is in the middle, and the 5 is to its left. If they are in the top row, then the 1 would have to be in the top row to make it add up to 10, and that doesn't work since the only top-row slot left is the top right. So the 4 and 5 are on the bottom row, and the 2 is in the bottom right. So the 3 is in top right, 1 is top left, and that leaves 6 in top middle and 2 in bottom right. Solution: 163 542 115: Since AB + BA is going to be a multiple of 11, the answer has to be A, and the numbers are 13 + 31. 008: 1) There's a circle in the oval shape in the top picture, making it look VERY eye-like. The "pupil" circle is gone from the bottom picture. 2) The chimney is straight in the bottom picture, but bent in the top one. 3) There's an extra branch to the left of the house in the bottom picture, but it's missing from the top picture. Also, the bottom picture is cropped more than the top one, or taken with a bit of a zoom lens, because the bottom of the brick border isn't visible -- but I bet that's not one of the differences the game is looking for.
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May 18, 2016 14:41
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129: This is a "cross every bridge twice" problem, so we're looking for the odd-numbered junctions. C has three branches, and another unlabeled junction just above F has three. So the mole can accomplish his task, but only if he starts from C. 060: Had to look at the second hint before I got it. Just draw a circle of any size around the statue. Say, through the middle of the doughnut. The outer doughnut and the inner doughnut now have the same shape (though not the same size). 085: I'm not actually getting this one. 076: It's the tower itself. 005: Big oval mirror just left of the doorway (cracked in bottom photo). Wine bottle at bottom center (spilling wine in top photo). Haven't found the third yet, and I think I'm not going to. There's a plank line visible in the top picture under the wine spill, which is missing in the bottom picture, but I'm guessing the game will lump that in with the wine spill as just one difference, right? 119: There's a way to do it in only ONE weighing. Open bags 1 through 4 and put their coins in four separate piles. Then take 1 coin from pile 1, 2 coins from pile 2, 3 coins from pile 3, and 4 coins from pile 4, and put them all into bag 1. If all of the coins from bags 1 through 4 are genuine, then that will weigh 100 grams, and bag 5 (by elimination) must contain the fake coins. Otherwise you'll be short by N grams, and N will be the number of the fake-coin bag. I.e., if that collection weighs 98 grams, then it had two fake coins in it, so they came from bag 2. Now, you could have done that by also adding 5 coins from bag 5, for a total expected weight of 150 grams that's guaranteed to be short. But even if the scale had a max of 100 grams, the approach above would have allowed for only one weighing to solve the puzzle.
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May 22, 2016 01:23
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110: "Didn't walk any farther than necessary" would imply that he never doubled back; he only headed east or south at intersections. The only route I've found that passes in front of one hat shop, one flower shop, and turns at one (and only one) cafe is ESSEESE. 111: To end up at your starting point without making any U-turns, you'd have to have an even number of turns. The two who reported an odd number of turns, 105 and 113, are suspicious. 092: There's an obvious choice in the NE corner with two silver and three copper, worth 9. To beat that, we'd need 10 or more from somewhere. There's no way to get two golds in five adjacent squares, but there is a score of 10 in the lower right-hand corner: take squares F3, E3, E2, E1 and D2. That gives a gold, a silver and two copper, for a total of 10. There's no way to get 11, so that's the best. 064: If you pull out 100 stones, there's an almost-50% chance that the one left over will be black, and a just-over-50% chance that the one left over will be white. If a white stone is left over, you just won 100 gold. If a black stone is left over, you won nothing. But since any other number of stones pulled out has roughly similar chances of being half-white and half-black (e.g., pulling out just 2 stones has roughly 50% chances of winning), that means you're best off pulling out 100 stones. 071: When he was her current age, she was 10. This works if they're five years apart, so he's now 20 and she is now 15. 069: Three passes. On the first one, copy E and F, the "front" and "back" covers. On the second one, fold D inside and photocopy it with A, the page that's opposite F. (You can tell which it is by where the string is). On the third pass, photocopy B and C together. 096: D has only four fingers, so it's out. The tail on A is too long, so it's out. And C's arms are missing the "shoulder", so it's out. B is the correct answer. 120: If D+A = F+H+E, by elimination in round one B+C = G. So from round two, A=H. So we can remove A and H from round three's equation and see that D=F+E. So the answer is that D alone should compete against E and F if you want a draw. 075: There are five each of three kinds of sweets, and ten of the fourth kind. So we can draw lines between any touching pair of the "unique" kinds of sweets, and from there the solution is pretty easy to find. 074: What kind of bottle is THAT?!? There is no path leading from an opening to the flowers, and all the openings connect to one or the other of the garlic cloves. However, openings 1 and 3 lead to the leftmost garlic clove, and opening 2 leads to the rightmost clove. So if you seal opening 2 you cut off the smell of the rightmost clove, and if you seal opening 1 you REDUCE the smell of the leftmost clove since it takes further for its smell to reach opening 2. I see two possible answers: BEST: put the corks right down inside the bottle, where the garlic smell emerges from the bulbs. But that's probably not legal. So... SECOND-BEST: Seal the leftmost and middle openings. If all you have is two corks, that's the best you'll be able to do to mitigate the smell.
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May 24, 2016 07:02
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098: I solved this before reading down to see that it was solved within the LP. Swap positions 2 and 5 and now there are three jars on the left, and three cans on the right. Done. ... Wait. Is THAT what they meant? Because "you must always move two containers at once" is easily fulfillable with the solution I thought of: pick one up with your left hand and one with your right hand, and swap them. Voila, you've moved two containers at once. Bad wording on that puzzle. 113: On the middle die, the top and bottom can't be 1/6 or 2/5 since you couldn't add up the touching faces to 5 with a 5 or a 6 in there. If the bottom face of the middle die was 4, the top face of the bottom die would be 1, but we can see the 1 and it's not on top. So the bottom face of the middle die is 3, and its top face is 4. Therefore the bottom face of the top die is 1, and its top face is 6. 002: You know what, I don't enjoy these, so I'm going to skip this one. 003: Ditto. 124: Looks like C can make all those shapes. 086: The figures are A,B,C,D,E,F in seven-letter segment displays, mirrored. So figure 1 should look like figure 2 minus the two bottom horizontal segments.
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May 28, 2016 08:10
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112: There's actually nothing in the wording of this puzzle that says the light goes out if you don't hold any of those contact points together... so I would draw one wire between the light's second terminal and the battery's negative terminal. Ignore A, B, and C: they're red herrings. 065: I'll start by trying a 6 next to the 4 so that the sum is 10; if that works, there's no higher possible sum. That leaves the leftmost line with 5, 3 and 2 to add up to 10, leaving 1 for the middle spot. Therefore 5 is at the bottom, 3 in the left middle, and 2 in the left top. 073: The first stick is tied way on the left rather than in the middle, so the left half must be heavier; E through H are out. The second stick is also tied on the left, so its left side must be heavier, and that's can A, which is filled with water. The other sticks are all tired roughly in the middle (except for the one that balances B+C with D), as befits sticks with equal weights on both sides.
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Jun 3, 2016 14:55
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W18: I feel like we saw this one before. Whether we did or not, this is a "cross each bridge once only" puzzle in 3-D space. Adding a wire at either A or C would reduce the problem down to only 2 odd-numbered vertices, but if we add the wire at A, the ant would be starting on an even-numbered vertex, while there would still be two odd-numbered vertices available. So we should add a wire at C to reduce the odd-numbered vertices to 2 while still letting the ant start at one of those two. W19: If Bill's age times two equals Rupert's age times three, Bill must have an age divisible by three and Rupert's must be divisible by two. Since we know both boys are under ten, it's pretty quick to check Bill's age against 3, 6 and 9 and see which one works. Bill is 9 and Rupert is 6. W20: Since there would have to be exactly four parts colored green, D must be the right answer. W21: The formula contains (x-x) as one of the multiplication terms. (x-x) must be 0, so the whole formula must equal zero. W22: The only way those locations work out is if there are five rows and five columns, so there are 25 total gems.
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Jun 9, 2016 13:02
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Photo 001: 1) The lamp by the window is lit in the first picture, but off in the second picture. 2) The number of bottles under the mirror has changed (from two to one). 3) The doorknob is on the opposite side. 114: Um? Is this supposed to be a trick question? The obvious answer is that A would receive 5 coins and B would receive 4, a rate of one coin per day. Why this would be worth 40 picarats is beyond me. If you're supposed to conclude that A worked slower than B, and therefore B actually did more work than A... well, that's not at all implied by the question. So I'm going with "This is a bad puzzle." If it's the obvious answer, it's too simple. And if it's NOT the obvious answer (as the hint coin implies), then the answer is probably that B receives 5 coins and A receives 4... but there's not enough information in the puzzle to figure that out. 109: No need to add a series of fractions here. The daughter had a 10-meter head start, and her dad is catching up to her at a net 1 m/s, so it will take him 10 seconds to catch up. The dog is running the whole time at a rate of 5 m/s, so the dog will run 50 meters total. 090: Since the middle spot has a value of 10, J is useless since it has a theoretical value of 10 but will never fire. Photo 004: 1) There's one document missing from the display case in the lower right. 2) An extra island has appeared in the map behind the podium that's more or less in the center of the image. 3) The white cuff on the jacket (just behind the painting) has become black. 095: In theory, there's enough volume for six books, but you'd have to cut one in half to make that fit. But here's a layout that fits 5 books into the box (X marks an empty space): Bottom layer: AAB AAB CCX Second layer: DDB DDB CCX Top layer: EEX EEX XXX But in this layout, book E can slide around inside the box if it gets moved. If we instead flip books D and E so that they're spine-up, we have a layout that not only packs as many books as possible into the box, but also ensures that your books won't slide around inside the box while it's in the moving truck (I mean, why else would you be packing books into a box?): Bottom layer: BBA BBA CCX Second layer: DEA DEA CCX Third layer: DEX DEX XXX
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Jul 18, 2016 23:44
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087: DHCS SCHD HDSC CSDH Photo 006: 1. No bush by the right side of the entrance. 2. Pickaxe handle missing from cart in lower right. 3. Circle (wooden log seen end-on) missing from lowest point on left side of entrance. 066: The description isn't very clear, but it seems to be saying that when you turn the red dial clockwise, the blue one turns counterclockwise. In that case, turn the red dial to 2, 7, 1, and 4 in that order. 079: Towards B. 078: Using chess notation where the entrace is A1 and the exit is C3, the key door is the one between B1 and B2. If it opened the other way, you could go A1-B1-B2-A2-A3-B3-C3-C2-C1, then go back to B1 and do the whole sequence between B1 and C3 again -- only this time go out the exit. The puzzle never said that you can only go through each room ONCE... 081: Ooh, Mastermind! I'll label the rows A through D from top to bottom, so I can talk about rows without getting mixed up with numbers. Looking at B and D, it's clear that 3 and 4 can't BOTH be correct since D only had one orange light. So the 5 in B had to have been correct, since moving it in D turned off one orange light and turned on one white light. Now looking at row C, I'm guessing on instinct that its 1 is correct, which gives us .15. and the code must be (looking at A) 0154. Checking... yes, 0154 is correct.
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Jul 29, 2016 12:03
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082: The positive and negative terminals are: -.+. +..- Connect top 1 to bottom 4, and top 3 to bottom 1. 077: Looking at the weights, clearly the crescent moon is the light one, the sphere is the medium one, and the star is the heavy one. That means the weights from left to right right now are: 7, 4, 5, 3, 7 So we want to add a weight of 1, which is the light ornament: the crescent moon shape. 121: You'll need at least one lamp at the end of the three short diagonal roads. For best efficiency, they should be at the top of the leftmost diagonal road, at the bottom of the middle diagonal road, and at the bottom of the rightmost diagonal road. That covers all the vertical and horizontal roads except for the leftmost vertical and topmost horizontal, and adding one more lamp at the upper-left corner will take care of that. So in chess notation with A1 being the lower-left intersection and D4 being the upper-right, place lamps at: A4, B3, D1, C2. Out of time for now; I'll come back and finish the other puzzles in a few hours. 122: Using chess notation, A1 is the lower-left intersection and E5 is the upper-right intersection. A5 and B5 and B4 aren't really intersections, but this keeps things consistent. The lower-right diagonal road goes between D2 and E1; a lamp at D2 would be partly redundant with the C2-D3 road's lamp, so we'll put this one at E1. The leftmost diagonal is most efficient if placed at B2 rather than C3, since that takes care of the short vertical road in the B column, which means that the C2-D3 road's lamp should go on D3. Lamps so far: B2, D3, E1. Rows 4 and 5 still need coverage, and columns A and C still need coverage. C5 will cover C and 5 completely, and A4 will cover A and 4 completely, so we're done. Place lamps at: A4, B2, C5, D3, E1. Photo no. 7: skipping. 097: Got back to it. Here's a link to my solution: http://lpix.org/2527568/097-solved.png 125: A good rule of thumb for each step is to look at each adjacent clearing and see how many unvisited clearings it connects to (the one you're standing in is clearly visited). If it only connects to one unvisited clearing, then that's where you HAVE to go next or you'll never be able to get back to it without visiting at least one clearing twice. So the first step we should take is the clearing just to the east of the starting point. Then north, and now we have an actual choice. Also, there's a forced path right in the middle of the puzzle. Starting with the clearing that's North-then-North from the start point, the forced path goes N, N, W, W-ish, N-ish, NE. At this point we're right next to the exit, so if we've visited all the other clearings, we can leave. But that's not likely, for reasons I can't manage to explain clearly in text. More likely is that we'll have to go S, Se, SE, then N to the exit. So once we step onto the clearing that's N, N from the start, we now have a forced path to the exit. We therefore must find a path from the start to that N,N clearing that doesn't touch any of the clearings along the forced path. And with that constraint, the puzzle becomes solvable by eye. First half of the path, from the start point: Go E, N, N, SW, NW, SW, N, NE, SE. Now we're at the start of that forced path. From here, go: N, N, W, W-ish, N-ish, NE, S, SE, SE, N to the exit. Tax Refund fucked around with this message at 13:33 on Aug 4, 2016 |
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Aug 3, 2016 00:49
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123: This time the grid system is A1-F6, where A1 doesn't even exist: the road goes from A2 to B2 to B1. There are four diagonal roads: A5-B4, B3-C2, C4-D5, and D3-E2. If we place a lamp at B3, that forces lamps at A5 and E2, and the A5 lamp then forces a lamp at C4. If we place a lamp at C2, that forces lamps at D3 and D5, which is less efficient. So we'll go with B3, A5, E2 and C4 for our first lamps. To cover row 1, we need D1, and that leaves F6 to cover the last two roads. Solution: A5, B3, C4, D1, E2, F6.
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Aug 5, 2016 00:43
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126: The real-life answer would be: N, then E, then N again, then grab the trees for support while you work around to the exit that's RIGHT NEXT TO YOU. But the puzzle wants us to answer N, E, S, W, N, E, N. 127: Oh boy, is that convoluted. N, W, N to get to the upper-left corner. Then E, S, E, S, E to get to the lower-right corner. Then N, W, S, E, S, E to circle around to the spot where we can finally reach the exit, and finally N to the exit. 128: N, E, N, W, S, W, S, E, S, E, N, W, N 100: Go forward 3 planks with the right foot. Then 3 with the left foot. Then 1 R, 1 L, 3 R. Now at this point, the next steps could be 3 L, 1 R, 3 L, but that would land us one step beyond the red footprint, and I bet the puzzle won't accept that as a correct answer (even though it should be correct). So instead of that, we'll start over from the beginning and take a different approach. If we start out with 1 L, 1 R, 3 L, 3 R, 3 L, we'll be able to hit the red footprint exactly. The next steps are 1 R, 1 L, 3 R, 1 L and we're done.
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Aug 9, 2016 01:49
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130: I would circle the skirts on the left side of the picture. They seem to indicate that this photo was taken with the camera at ground level, angled up so that the chandelier appears right behind the dancing couple. 131: Only the single post in the 3 o'clock position is "inside" the looped rope. If it is pulled up, the man will be able to run away. 134: Following the pipe from 2, go "east" at the first junction, and you'll be following a pipe section with three valves. Turn those three valves and you'll have opened a path to the middle part of the boiler, without opening a steam path to either 1 or 3. Photo #9: 1. Lever next to the tracks (right by the big hole) is in a different position. 2. Section of railing near the bottom is bent vs. straight. Haven't found the third one yet. 088: If we aren't confined to the purple path, and the puzzle never says that we are, then a 1-space jump to the right, then 2 R, then 3 D, 1 D, 2 D, 3 D solves the puzzle in 6 jumps.
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Aug 15, 2016 07:09
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135: Well, this is easy. There's already a diagonal filled in, so we know the total is 8 + 3 + 6 + 1 = 18. So now we just fill in the remaining numbers by elimination, starting with the ones where three of the four numbers are placed, and we get: 8235 5328 1764 4671 The only tricky part was where to place the 8 when there were four open spots remaining, but it obviously needed to go in the row and column with the lowest values so far. 136: If you number the squares in chess notation with A1 at the bottom left of the board and B1 immediately to its right, then the hidden door is in square E3: it's where the path is pointing to. 132: The total price of the paintings is 275K. For the older brother to get twice the value of the younger brother's paintings, the four paintings they get must have a total price divisible by 3. So we need to find which number we can subtract to get a price divisible by 3, and we'll know the painting the valuer got. (There might be multiple candidates, so we might need to narrow it down further). Subtracting 20K gives 255K, divisible by 3. Subtracting 60K gives 215K, not divisible by 3. Subtracting 55K gives 220K, not divisible by 3. Subtracting 45K gives 230K, not divisible by 3. Subtracting 95K gives 180K, divisible by 3. So either the valuer got A or E. If he got A, there's no way to divide up the paintings evenly because 255 = 85 + 170 and there's no way to make 85 from the remaining paintings. But if the valuer got E, then the younger brother would need to get 60K of paintings and the older brother 120K, and that's easy: give the younger brother B, and the older one gets A, C and D. So the valuer got painting E, the most expensive one. Someone I suspect that he may have known the terms of the will, and arranged the prices in such a way that giving him E was the only way to fulfill the terms. If I were the brothers, I'd get in a second opinion. W23: We'll assume that neither child is adopted or the child of an earlier marriage, so both parents are older than both children. So if statements A and C are both true, then Dan must be one of the parents since two people are younger than him. But then statement D would be false, which would mean that Lisa and Sam would be the parents, leading to a contradiction. So at least one of A and C is false. Now let's look at A and B since they would let us conclude something about Mary. If B is true, then Mary is the mother, and so D must also be true otherwise we'd have a contradiction again. Therefore A and C would be false: Mary is older then Dan, and Sam is older than Dan. This all checks out (Mary and Sam are the parents), so "B is true" is probably a correct hypothesis. To prove it, let's look at "B is false". Then Mary is younger than Lisa, and that means Lisa is the mother. The other statement about Lisa is D; if we assume D is false, then A and C would be true. If so, then Sam is the father (from false D), but Sam is younger than Dan (from true C), so that's a contradiction. Therefore in the "B is false" hypothesis, D must be true, and since we know Lisa is the mother and she and Sam aren't a couple, that means Dan must be the father. Therefore Mary and Sam are the kids, and A and C must both be true. But that leaves us with three true statements (A, C and D), and that contradicts the puzzle terms. Therefore the "B is true" hypothesis is the only remaining possibility, which means that Mary and Sam are married to each other, Lisa is their daughter, and Dan is their son. W24: This one really IS easier with algebra, unlike some of the word problems. 9x + 7 = 7x + 9, so 9x - 7x = 9 - 7, and the mysterious number is 1 (one). W25: Judy would never claim a false name, so woman B cannot be Judy. Judy would also never claim that someone else is her, so woman A cannot be Judy. Woman C is Judy, therefore woman B is Anna, and woman A must be Ellie. 137: The sword in the hand of the knight on the right must be the real one. The knight on the left is also holding a weapon, but it's not a sword.
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Aug 19, 2016 11:59
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Dragonatrix posted:Hey so today's update is I think the biggest one of all which isn't much of a surprise, but it's also very GIF heavy so for the sake of the thread not choking itself to death I'll just test-post this one as well. Thought #1: The world is an illusion? Huh -- I didn't know Layton was Buddhist. Thought #2: No, no, Sofia, you're doing it wrong. "Guess what, honey? You're going to be a father!" THAT'S how you do it, not some enigmatic designed-to-be-misunderstood nonsense straight out of a Three's Company episode.
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Aug 21, 2016 09:42
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W26: Hide in the cage that the lion just left. Since the puzzle tells us that you forgot to lock it, that means that you could have locked it -- you have the keys. So just in case you forgot to lock any of the OTHER cages, or the other lions that got left out might still be roaming the zoo, it's safest to go into the now-empty lion cage and lock yourself in so no animals can get to you. W27: Looking at the center of the pattern and the little "bullet-shaped" things diagonally around the center square, A doesn't match. Then if you look at the triangles around the outer ring of the pattern, and specifically at the one in the SE position of the black-and-white pattern (the SW position of the colored patterns), C doesn't match. So the matching pattern is B. W28: First card to place is in square B3, which has to be a diamond since it shares a diagonal, row or column with the three other suits. Then it's easy enough to do the rest. HSCD CDHS DCSH SHDC W29: Since E is the only odd-arity intersection, that's the obvious starting point. The ending point will be the arity-3 intersection two north of E (and one south of C).
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Aug 22, 2016 11:24
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| # ¿ May 21, 2024 03:28 |
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W30: Third line gives us all the digits in the code, so there are no 4's or 5's anywhere in it. So the other three lines become: 2.3., one correct and one misplaced. 30.., one correct and one misplaced. ..02, one correct and one misplaced. Finally, 3102 also has one correct, and three misplaced. If 3 is correct in position 1 as lines two and three have it, then 0 can't be well-placed in positions 2 or 3, so it would have to be in position 4. But then line four could not have any correct placements in it. Therefore, 3 is MISPLACED in lines 2 and 3, which means that the 0 placement in line 2 is CORRECT. Therefore, line 4's placement of 2 must be correct. Therefore, line 1's placement of 3 must be correct. And now we have the code: 1032 W31: Um... what? The smallest roll you can make is 1 (on die A) plus 3 (on die B), totalling 4. Why would the hint say that this puzzle is "a bit mean"? If you're supposed to say "Hey wait, if I just roll die A twice then I could get 1 + 1"... well, the example given in the puzzle wouldn't work then, because if you are allowed to roll die A twice then you could get 5 + 5 = 10, but the puzzle said the highest number you could get would be nine. And the puzzle said that you can see all the die faces in the image, so it's not like it can pull a "well, there was another face on die B that we didn't show you, and that face had no dots on it at all" trick. And any "those are dots, not numbers" trick would also be ineligible, because the puzzle used the exact same "numbers" phrasing for the "highest number is nine" example. So I'm going to go with the obvious answer and say "Don't overthink this one; the obvious answer, 4, is correct." W32: Gonna need to turn that 6 around into a 9, right? Then you can get 43 x 3 = 129. W33: Tracing back from the exit seems easier to follow. Also, I'm going to call the north/south roads A through G, from left to right; and the east/west roads will be 1 through 6 from south to north. To get out, the bus can either turn east at E5 (it would have been heading north on E before), or at D5 (it would have been heading south on D before), or at B5 (it would have been heading north on B before). We can reject D5 right away as there's no way for the bus to turn south at D6, so that leaves E5 or B5. B5 looks very plausible -- I've found a route for it just by looking, so let's look at E5 and see if we can reject it. To head north on E, we would have had to come from E3, where we were previously going west. To get there, we would have had to be going south on F and then turned west at F3. The only way to be going south on F would be to turn south at F5, where we were previously going east. But wait: if we were already heading east on 5th street, there's no reason at all to turn south at F5! We would have just continued straight! Therefore, we can reject E5, and go with the route I saw at B5. In REVERSE order, it goes OUT-B5-B1-D1-D2-G2-G6-C6-IN. Therefore, the shortest route for the bus to take to the goal is IN-C6-G6-G2-D2-D1-B1-B5-OUT. You could add a little F5-F3-E3-E5 loop in there if you want, but then it wouldn't be the shortest route any longer. And that's it. Thank you for doing this: I had loads of fun solving most of these puzzles. ("Most" because one or two were badly written and/or missed a better solution, but out of over 150 puzzles, that's an excellent ratio).
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Aug 24, 2016 12:14
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