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Zaroff posted:Maybe I’m missing something, but can’t you solve #72 without any logic whatsoever? That's just a different kind of logic.
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Sep 14, 2019 00:14
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| # ? May 4, 2024 04:51 |
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 Featuring Chapter 8. Like, all of it. It's a relatively short chapter, all things considered.
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Sep 16, 2019 00:32
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076: 5 people. 077: Suppose they try to meet at 11:00. The narrator thinks his watch is 5 minutes slow, so he'll try to arrive when his watch says "10:55". But, it's actually 5 minutes fast, so it's really 10:50. Meanwhile, the narrator's friend will try to arrive at "11:05" (because he thinks his watch is 5 minutes fast). But that watch is actually 5 minutes slow, so he'll arrive at 11:10. That's 20 minutes after 10:50. 078: Let's say the rabbit jumps 70 cm, then spins while jumping so he lands with his tail pointing in the direction his head was pointing - so he jumps 30 cm further, for a total of 1m in the direction he wanted to go. Repeat this 10 times and he's traveled 10 meters in 20 jumps - 10 forward with a spin, 10 in the direction of his tail (which is pointing the same way). 079: A, on the left, is the smallest, and finishes fourth. As the second-fastest swimmer is standing next to B, they must be C. E, on the right, clearly has the fanciest outfit, but is slower than B - and B claims he isn't going to make first either. The only person left is D to claim first place. Followed by C, B, A, and E, in that order. 080: I give up! 145: 20 times. 01:26, 02:16, 06:12, 06:21, 10:26, 12:06, 16:02, 16:20 blaze it, 20:16, 21:06... and then 01:29, 02:19, 09:12, 09:21, 10:29, 12:09, 19:02, 19:20, 20:19, and 21:09. 081: Change the 6 on the right side to a 0. You can use the 9 on the left for a 6 if need be.
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Sep 16, 2019 01:47
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076: There are at least three people, since Guinevere and Nimue plus King Arthur makes three, but if there are only three then they'll have to be adjacent to one another. Adding a fourth person would make them sit opposite each other(since the people in attendance are sitting in evenly spaced intervals), so there must be five people at the table. 077: The person who thinks their watch is slow will arrive ten minutes before the appointed time, and the person who thinks their watch is fast will arrive ten minutes after the appointed time, which makes a difference of 20 minutes. 078: Jump forwards 70 cm, turn around in midair, hop another 30 cm facing backward, turn to face forward again. Two jumps ten times for a total of twenty jumps. 079: A is the smallest, so A will finish fourth. The second-fastest swimmer is standing next to B, so C is the second-fastest. E has the fanciest costume and so is slower than B, who is also not the fastest, which leaves D as the fastest swimmer. 080: Three people standing in a circle: a man in front of a male child in front of a woman in front of the man. 081: As it stands, the dice can't represent 07 or 08, so there needs to be a 0 on the right die. The 9 on the left die can also function as a 6, so changing the 6 on the right die will let you display all days of the month. 145: You can use the flip-six-to-make-nine trick here, too. The valid hours are 01, 02, 06, 09, 10, 12, 16, 19, 20, and 21. Since the 6/9 card can never be in the tens place of the minutes and the other three always can, every hour has two possible minute combinations for a total of 20 possible times.
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Sep 16, 2019 01:48
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#076 - It would seem 5 (none are opposite, and the other three are men). But I feel like there's something else going on here. #077- It's a little unclear what 'checking our own watches' means here. Does this mean 'going by the time on the watch' or 'going by what we think the time is, based on the watch and ignoring any other reasoning'? I assume the latter. If the speaker thinks their watch is slow, they think that it is actually five minutes later than the time displayed. If the watch reads 11:55, they think it is actually noon. When the watch does display 11:55, it is in fact 11:50 by world time. For the other person, if they read 12:05, they think it is actually noon. But since the watch is slow, it is in the same world reference frame 12:10 when it displays 12:05. So for a given meeting time, the person will arrive 20 minutes before their friend. #078 - 15. There's no indication that the jump tailward must take place immediately after the jump forward. Thus the rabbit can make 15 70 cm jumps forward, which will 'cover' the 10 m, and make tailward jumps later at its leisure. #079 - D. By S1, A or C must be #2. S2 (with S1) means B must be #3 or higher. S3 means the 'fancy' costume (E in the two-piece, I'd guess) must be #4 or #5, and B cannot be #5. S4 puts A as the smallest at #4, which puts E at #5. S5 is meaningless, as the winner must beat everyone, not just those beside them [I know they probably meant 'come in one place ahead of']. However, we already have to put B at #3, and C at #2, so D is #1. #80 - 3. If we allow circular positioning, then (-> showing facing) : Female-> Male Child-> Male-> ... works. If not, then <-Female [Child] <-Male [Child] Male [Child]-> satisfies it. e: There are multiple patterns of children that satisfy the conditions; a female child and one male child, or two male children. #145 - 27 if I counted right. The cards can create the digits 0,1,2/7, and 6/9 (I have literally seen a clock using almost exactly this number style, so the game better not be messing with us here). The times that can be displayed are: 0126,0129,0216,0219,0612,0617,0621,0716,0719, 0912,0917,0921,1026,1029,1216,1219,1602,1607, 1620,1706,1709,1902,1920,2016,2019,2106,2109 #081 - Change the 6 on the right to a 0, as the 9 can be flipped and you need to be able to write 7-9 with the right cube as 0. Kangra fucked around with this message at 06:03 on Sep 16, 2019 |
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Sep 16, 2019 05:47
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76. So we know there must be at least 4, or one of the two would be adjacent. However, that means they would be opposite each other, so adding one more person, to 5, fixes that. 77. After I adjust for my watch, I'm 10 minutes early. Friendo, correspondingly, is 10 minutes late after he adjusts. Since they both adjust in different directions, it's 20 minutes. 78: If the rabbit spins in midair to face the opposite way (so its second jump is along the direction of its tail, or "forward"), it goes 70-30. So that would be 20 jumps. 79: By clue 1, it's not A or C. By clue 2, it's not B. However, we know A comes in 4th by clue 4, so C must be second. Thus it's D, who beats C to win by clue 5. 80: So we have M<C<F, at least, by the first and last clues. I think we can go MCMF, with the first male facing left (one male and female standing behind a male), child facing left (one male in front, one female behind), and the female facing right (two males behind a female). Unless of course the child's gender can count into it, in which case I think we can manage 3 with a male child. 81: 07 and 08 can't be managed as is, so the 6 on the right die needs to become a 0. That technically means 06 is no longer possible, but that's what inverting 9 is for. 145: We have 6-9 as a possibility. So by 24-hour clock, valid hours are 01, 02, 06, 09, 10, 12, 16, 19, 20, 21. Neither 6 nor 9 can be in the tens place for minutes. So for 06/09/16/19, we have 8 options, and for 01/02/10/12/20/21, we have 12 options (01:26 versus 01:29, for example). This leaves us with a total of 20.
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Sep 16, 2019 06:05
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076: We know at least two women were present. If there were only 2 guests, Guinevere and Nimue were opposite each other. If there were 3, they were adjacent. If there were 4, they were either adjacent or opposite. If there were 5 guests, they could be seated man-woman-man-woman-man, and nobody would be directly opposite anyone else if five people were evenly spaced around a round table. The answer is 5. 077: Someone who thinks their watch is five minutes slow is going to take that into account, and show up for a 2:00 meeting when their watch shows 1:55. Someone who thinks their watch is five minutes fast will show up for that same meeting when their watch shows 2:05. The first guy's watch is really five minutes fast, so when it shows 1:55 and he turns up for the meeting, it's really 1:50. The second guy's watch is really five minutes slow, so when he shows up, his watch says 2:05 but it's really 2:10. So the answer is that the narrator will show up 20 minutes before his friend (no minus sign needed). 078: Nothing was ever said about not turning around 180° during the jump! So after a 70 cm jump forwards, the rabbit could land facing backwards, then do a 30 cm jump "in the direction of its tail" while still making progress towards the goal, and turning around again at the same time. It is, after all, an "uncommonly clever" rabbit. So each pair of jumps gets it 1 m towards the goal, so it takes 20 jumps to go 10 m. 079: We can match two statements to their speakers already: A is the smallest who will finish fourth, and B has the biggest body but will not win. From the first statement, C is therefore the second-fastest swimmer. The last two statements belong to someone who is slower than B, and someone who's confident they will win. Since that person says they'll "beat the person beside them" to win, they're beside C, so D will win. 080: If you have people stand in a circle, then the minimum is 3 people: a man, a woman, and a boy. All of them can be considered "in front of" or "behind" the other people in the circle, depending on how you trace it. (Though three people standing in a circle doesn't *really* work, IMHO, but I bet that's the answer the puzzle is looking for). 145: With these fonts, it doesn't look like the 2 can be turned around to become a 5, so the only dual-use card would be the 6 (becoming a 9, of course). Neither 6 nor 9 can start the minutes or the hours section, which eliminates many of the possible combinations. So there's 01:26, 01:29, 02:16, 02:19, 06:12, 06:21, 09:12, 09:21, 10:26, 10:29, 12:06, 12:09, 16:02, 16:20, 19:02, 19:20, 20:16, 20:19, 21:06, and 21:09. That's a total of 20 different valid combinations. 081: As-is, the dice can't display the date 09. Change the 6 to a 0, and the 9 can be used upside-down as a 6. Then either die can be used as the first digit 0, 1 or 2, and the other digits 3-9 are all representable as final digits on one die or the other. Plus you can still make 30 and 31.
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Sep 16, 2019 09:13
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Kangra posted:#145 - 27 if I counted right. The cards can create the digits 0,1,2/7, and 6/9 (I have literally seen a clock using almost exactly this number style, so the game better not be messing with us here). The times that can be displayed are: Now that I think about it, I've seen Upside-Down 2 as 7 on clocks before, but I don't know if the puzzle maker has...
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Sep 16, 2019 19:54
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Y'know, for Chinatown, there's not many Chinese people about. Although maybe that's for the best considering the game's artstyle.
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Sep 17, 2019 00:31
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Kangra posted:#145 - 27 if I counted right. The cards can create the digits 0,1,2/7, and 6/9 (I have literally seen a clock using almost exactly this number style, so the game better not be messing with us here). The times that can be displayed are: The second hint for this puzzle says "There's one card that can actually display two different numbers!". Just one card, not two, so it seems they're only counting 6/9 as a valid flip, and not counting 2/7 as a valid flip.
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Sep 17, 2019 14:43
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 Back to Chinatown we go. Again. It probably sounds more repetitive than it is, really.
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Sep 18, 2019 01:36
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Nahh, I remember it being repetitive too. 082: B is smaller than the other two and filling up C seems kinda difficult without packing the flour so tightly that it doesn't fall out when you tip the halves of the boxes sideways to put them together. Flipping A upside-down and filling it normally is probably the way to go. 084: One minute for A to row B across + one minute for B to row back + one minute for C to row D across + one minute for D to row back + two minutes for B and D to row across = six minutes, unless I'm missing some other trick. 085: 087: Ahh, good ol' algebra. 12 = 2J+2P = 3S+3P = 6J+6S. Therefore, J+P = 6, S+P = 4, and J+S = 2. S = J-2, P = J+2. Substitute in and solve: 12 = 2J+2(J+2) = 4J+4, so J = 2, which means S = 0. Sam doesn't get an allowance! 088: No need for equations with this one. Since the total involves a half-pound, we know that either the £1 or £3 item was bought at least three times. Can't be the £3 item; that's £7.50 of stuff and there's three other items. That means £2.50 pounds in £1 items, plus three other items totaling £7, which would be two £2 items and one £3 item. tl;dr Three £1, two £2, and one £3 item(s). 089: The fourth strip should be the one on the left side of the photo—notice how none of the strips show the head of the white bird from the fourth strip? That means H and I are in first and second place. The red bird matches up with the red tail in the third strip, so the third-place winner is F. Edit: Wrong letter in 082. Bloody Emissary fucked around with this message at 06:05 on Sep 18, 2019 |
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Sep 18, 2019 03:20
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Cue ominous chords.
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Sep 18, 2019 03:29
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82: Box A. It's larger than B, and while it opens near the bottom, it can be flipped, so it should hold more than C. 84: 6 minutes. A and B cross with A rowing (1 min), B rows back (1 min), B and C cross with C rowing (1 min), one of the three rows back (2 min), and them and D cross with D rowing (1 min). 85: I'm thinking switch the rightmost two for a solution with 2 moves, but something about the way the second hint is worded is making me wonder. If moving only the pork in the noodles would work then maybe just one move, but I'm not sure about how that sounds. 87: Just needs some algebra to do, so 2J + 2P = 12 -> 2(J + P) = 12 -> J + P = 6 3S + 3P = 12 -> 3(S + P) = 12 -> S + P = 4 6J + 6S = 12 -> 6(J + S) = 12 -> J + S = 2 P = 6 - J and S = 2 - J 3(2 - J) + 3(6 - J) = 12 -> 6 - 3J + 18 - 3J = 12 -> -6J = -12 -> J = 2 So, Jeff gets £2 a week, Pete gets £4, and Sam gets nothing. 88: 3 £1, 2 £2, and 1 £3. Need to get the discount on either the £1 or £3 to get the .50, and doing that with the £3 means you couldn't get one of each of the others without going over £9.50. After that, the other two just need to add up to £7, so one needs to add up to an odd price. 89: F. Original picture would be 4-3-5-2-1 starting from the left.
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Sep 18, 2019 03:47
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082: Flip Box A upside down and it should work fine. 084: 1. A rows over with B in tow. A gets out. (1 min) 2. B rows back. B gets out and C and D get in. (1 min) 3. C rows D over, and C gets out. (1 min) 4. D rows back. B gets in. (1 min) 5. Either of B or D rows over, and both officers get out. (2 min) It takes 6 minutes total. 085: We're doing this again? Swap the positions of the two rightmost bowls and you're done. That's 2 moves. Alternately, if we can move the pork between bowls, move the pork between the two rightmost bowls. That's 1 move. 087: This is algebra. (P+J)*2 = 12, so P+J = 6 (P+S)*3 = 12, so P+S = 4 (J+S)*6 = 12, so J+S = 2 Add the first two equations together and we get 2P+J+S = 10. Then subtract the third and we get 2P = 8. So Pete gets £4 a week. With that nailed down, we can conclude that Jeff gets £2 a week... and Sam doesn't get any! Ooooof. 088: This is almost algebra. Since you have 50 pence on the price, you had to get a £1 or £3 item for half price. Suppose you bought three £3 items - that would be £7.50. You have £2 left on the bill and need to spend it on four items, which is impossible. So, you must have bought 3 £1 items, for a total of £2.50. From there, you need to pay £7 for three more items, which you can do by also buying two £2 items and a £3 item. So you bought 3x £1 items, 2x £2 items, and 1x £3 item. 089: F.
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Sep 18, 2019 05:05
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I should have known they wouldn't do the 2/7 thing. #082 - I imagine they're going for A here (inverting the box), but it's really impossible to tell. C is more likely to be a better box, since a lot of boxes with lids like that actually have the box portion as the full height, and the lid makes a better seal if it comes down farther. #084 - 6 minutes. It's easy to see that 5 trips is the minimum required to get them all across, so someone will have to row twice. If D+C cross (1'), C returns(1'),B+A cross (1'),B returns(1'), then B/C cross together (2'). #085 - 2 Switch the two on the right. #087 - [0,1) or 0 if an integer. This is a matter of determining the region defined by the constraints, essentially the feasible region of a linear programming problem. The only one fully constrained is the pooling of Sam and Jeff ("it would take" vs "could") but a reasonable assumption is that in the other cases the car can't be purchased any earlier than the week indicated. Pete's statement can be interpreted as 5S + 5J < 12 and 6S + 6J >= 12. With the same number of weeks is used for each person, it's easier to just divide the price by the number of weeks. The full set is: 6 <= P + J { < 12 implied} 4 <= S + P { < 6 implied} 2 <= S + J < 2.4 (additionally S,P,J all >= 0). Without the implicit constraints on Pete's income, Sam's pocket money is anywhere between 0 and just under £2.4 (£2.39, or £2 7 11 3/4 if we're in the past). Otherwise the solutions are in a pyramidal region with a rectangular base and an apex at (1,1,5).  But since we assume strictly less, that apex can't be a solution, so Sam's income is in the range [0,1). If it's constrained to be an integer number of pounds, then it must be 0. #088 - 3 apples (£2.50), 2 pumpkin(?) (£4), and 1 jar of chutney (£3). To get a .50, you have to purchase either a £3- or £1-item at discount. But if you purchase 3 jars, that's £7.50 with just three items, and you can't get six items as the cheapest is £1. Getting three of the cheapest items uses up £2.50, and you must buy more expensive items to reach £9.50 with six. #089 - F. This assumes the runners are moving forward, to the left in the photo.The cat in F's strip connects to the JK strip; the head of the bird near J connects to the DE strip, and the blue bird at the bottom connects to the ABC strip. So all of those are behind FG. HI is the frontmost, as can be seen by the rear of I's head in the FG strip, so the top three finishers are H,I, and F (ahead of G). Kangra fucked around with this message at 14:58 on Sep 19, 2019 |
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Sep 19, 2019 14:52
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082: Container A, once you turn it right-side-up, has the largest capacity in the body of the container. 084: First, A rows across while B is a passenger: 1 minute. Next, B rows the boat back: 1 minute. Then C rows across while D is a passenger: 1 minute. Then D rows the boat back: 1 minute. Then B and D row back across; no matter who rows, it's the second time for each of them, so 2 minutes. Total time: 6 minutes. 085: I had to look at the hints to get this one. While you could swap the two last bowls in two moves, you could also just pick up the pork from bowl 4 and deposit it into bowl 5, which would count as one move. 087: In algebra terms, J+P = 6, S+P = 4, and J+S = 2. Since S+P = 4, P is at most 4, assuming there are no negative numbers here. From J+P = 6, J is at least 2, and from J+S = 2, J is at most 2. So J = 2, and Sam actually gets 0 (zero) pounds per week! 088: Buying three ₤1 items will cost you ₤2.50. Buying three ₤2 items will cost you ₤5.00, and buying three ₤3 items will cost you ₤7.50. Since you spent ₤9.50, you can't have bought three ₤3 items (total ₤7.50) since you bought six items: it's impossible to spend just ₤2 on three items. So the 50 pence must have come from buying three ₤1 items for ₤2.50, and the remaining ₤7 would be ₤2, ₤2, ₤3. So you bought three ₤1 items, two ₤2 items, and one ₤3 item. 089: Strip 2 is directly in front of strip 1 (blue bird at bottom). Strip 3 is directly in front of strip 5 (cat and white bird at bottom). Strip 4 is directly in front of strip 3 (red bird at top). Strip 5 is directly in front of strip 2 (white bird at top). So the order of the strips from front to back is 4, 3, 5, 2, 1. So person F is in third place.
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Sep 20, 2019 01:18
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I have to admit, this game does have the best hook to it, with, y'know. "It's London, X years in the future, and the villain is... well..."
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Nov 22, 2019 02:30
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Heya, folks. I've had this pretty much completely ready since... September... I just didn't post it because, uh.. .  Better three months late than never?
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Dec 18, 2019 01:27
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90: Even at a glance, it's not C. The design can't align near the junction. A bit of looking and mental assembling later, B also looks like it runs into a junction that makes it impossible. So looks like A? 91: It took a bit of looking to notice that G is slightly misaligned. The line in the middle is directly perpendicular to the point of the star in the first image. 92:  That lets you get two identical squares, so long as you can rotate one. 93:  95: Because of the block on the bottom, it can't be C. The green block in the center means it can't be B. The pink block in the center means it can't be A. So that means it's D. Puzzle Battle: 4 vs 3, 3 vs 2 for the two wins; 1 vs 1 for the tie, 1 vs 5, 1 vs 4 for the two losses. This avoids defeat through a tie.
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Dec 18, 2019 02:34
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90: A. 91: G. The crossbar of the 'do not' sign is parallel to the edge of the image in the left image, and slightly askew in the right image. 92:  93:  95: D. Puzzle Battle: 1 vs 5 and 1 vs 4 (Dmitri wins 2), 4 vs 3 and 3 vs 2 (Layton wins 2), 1 vs 1 (tie). You don't get defeated exactly...
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Dec 18, 2019 09:57
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 Last update for the year, to round out this in-game chapter.
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Dec 24, 2019 00:10
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Wait ... what?
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Dec 24, 2019 00:59
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096: Start at the lower of the two stars in the second column from the left 151: 98 - 59, turning the rightmost square in the second row of the first number of one side white and the leftmost square of the fourth row of the second number of the other side green 097: Start by going down, there's no choices to make on the route aside from a really short dead end
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Dec 24, 2019 01:16
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It's worth noting that this is the only Layton game where you solve a puzzle as Don Paulo.
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Dec 24, 2019 02:31
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96: Not too hard. Start from the star in row 3, column 2. It's really just a matter of seeing when you run into a corner. 151: Turn it into 98 and 59. Brute forcing it isn't too bad. 97: Down, right, up, left, down, left, and it's obvious from there.
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Dec 24, 2019 03:15
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96:  151: 59 and 98. 97: 
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Dec 24, 2019 12:06
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Okay, so at this point it's really clear that we are not, in fact, in the future. I had my doubts before, but the don paolo reveal more or less spells it out. Every single person who we saw a future version of, has now been debunked: Future layton was not Layton Future Schrader/Dean are Paolo. Future Luke just got debunked by Layton, since yes, he clearly does not remember all this as already happening, so i'm about 100% sure he's ALSO an impostor of some sort. Side bet: he's actually Claire. For added proof, remember the ONE character we met in the future who didn't age a day: Pavel. With this in mind, it's rather clear that the clock shop is, in fact, not a time machine but a teleporter. That still leaves 2 questions: 1. Why build a replica of london elsewhere? 2. What's fake Luke's angle?
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Dec 24, 2019 17:19
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Yeah, I'm most suspicious of FLuke. I actually thought he would've been Don Paulo. The solution to #90 did a decent job of showing how you solve something like that: Match the sides that must connect. If you start at the places already joined (or almost joined) you can follow the edges and connect them to their closest partner, until you reach the ones that are supposed to match. #96 This one is pretty easy to solve algorithmically. Pick any square. As you follow the path, X out the squares you visit. If you cannot proceed farther, pick another square that hasn't been X'd out. If no squares remain to pick, then the one you chose is the answer. #151 It's sort of weird that they just picked a random number for the difference, instead of having it be 'find the largest' or something. I guess they wanted it to be harder to guess wrong? (Incidentally the largest difference seems to be 48 as far as I can tell.) The 5 can become 9, and 8 can become 9, making 98 - 59. #97 Reasonably easy to solve this one by going backwards; you can never make a turn at a 4-way connection, and can only go into a 3-way connection if you are turning into the base of the 'T'. If you start by heading south from the start, the only 'choice' you then have to make is at the one intersection in the center right, where the dead end is right there next to it.
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Dec 24, 2019 17:54
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Regallion posted:For added proof, remember the ONE character we met in the future who didn't age a day: Pavel. Hey, you're right! It would be like him to blunder into the plot while looking for his... OK, I've forgotten what exactly he was trying to find this time 😅
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Dec 24, 2019 22:34
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Quackles posted:Hey, you're right! It would be like him to blunder into the plot while looking for his... He was looking to explore subterranean caves beneath Rome.
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Dec 25, 2019 09:03
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I suppose it's a bit weird that we haven't had a seemingly requisite "suddenly a lot of puzzles happen" update. Well, now we have. 
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Jan 6, 2020 01:44
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98: Left, down 2, left 2, slide -> down, slide -> right until the banana -> left 3, up, slide -> navigate around all the bananas till the end. 99: Assuming you can't loop around, the bottom right door is the only possibility. 101: Starting with D, circle around the bottom right. Then do the same with A. Then B circles left, and then around. That leaves C for an easy finale. 102: 33+24+27+15. 103: Middle, bottom, then bottom left. 104: Looking at the first two statements, can't be A, because that'd contradict B. It can't be B, because that would make him a liar, and the liar is the cake eater. That leaves C and D. C's substantiated by D, so it has to be D. 105: A, because it has to be reversed. 143: I'll have to think about this one. 106: 5+3 = 8. The 8 can be made by reversing the 3.
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Jan 6, 2020 05:03
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98:  99: Bottom right. The only door with two doors below it and two to the right is the top left, so we move down and to the right and... 101:  102: 27 + 33 +15+ 24. 103:  104: It can't be D, because then D and C would be lying. So D is telling the truth, and so is C. If B was lying, then someone else would have had to have eaten the cake, which is contrary to what the puzzle tells us. So, A ate the cake (and then lied about it). 105: It has to be A. When you put your finger on something, your print is reversed from what appears on your fingertip! 143:  106: Flip the '5' stamp and stamp 1+1=2. Two stamps used.
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Jan 6, 2020 06:13
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104: The kid who had the cake was the one who ate it. He lied about saving it for later.
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Jan 6, 2020 06:29
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Quackles posted:143: I really should have figured it was that shape, given that's the one it is on my keyboard. I'm just so much more used to the digital version, and was trying for that instead.
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Jan 6, 2020 06:37
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 An eternal friendship is great and all, but it's no eternal diva.
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Jan 8, 2020 01:23
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131: When she bought the pen, it was 10% off £30, or £27. Then, her friend bought it off her for 110% of £27, or £29.70. So she made £2.70 on the deal. That's 270 pence. W05: X○X○X○X ▪️X▪️X▪️X▪️ X△X△X○X W08: Circle the woman's hair, poking out from under her hat and goggles.. W09: 12 * 21 = 252. It's a palindrome!
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Jan 8, 2020 02:27
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If you turn a 5 upside down, it's still a 5. To make a 2, it would have to be mirrored.
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Jan 8, 2020 06:19
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| # ? May 4, 2024 04:51 |
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fractalairduct posted:If you turn a 5 upside down, it's still a 5. To make a 2, it would have to be mirrored. I, uh, realize that now :P
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Jan 8, 2020 06:30
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