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Ooh, another Professor Layton LP! Here are my answers: 006: This is the Towers of Hanoi. The key is (REDACTED). With just three pieces, the moves go as follows (1 is the smallest piece, 3 is the largest, and plates are (L)eft, (M)iddle and (R)ight): 1->R, 2->M, 1->M, 3->R, 1->L, 2->R, 1->R. Total: 7 moves. 011: Mark the squares like a chessboard, with A1 at the bottom left, A4 at the top left, and D4 at the top right. Then one piece is A2, A1, B1, C1. Another piece is A4, A3, B3, B2. A third piece is B4, C4, C3, D4. The fourth and last piece is D3, D2, C2, D1. 010: 3 colors needed. All but the lower left-hand piece can be colored with just two, but it and the two to its right all touch each other, and three colors are needed for that part of the diagram. 008: The total price is 105 of whatever currency (picarats?) is used here, so Luke's order cost 70 picarats and Layton's cost 35. The only items that can add up to 35 picarats are 8, 13 and 14. So by elimination, Luke's order must have been the items that cost 12, 16, 17 and 25 picarats. 009: Design B cannot stand up to gravity. The two outside glasses on the second row have nothing holding them up and will fall left and right. Design A is dangerous and would tip over too easily to be wise, but it's not impossible, while design B is totally impossible. 013: I could draw a geometric diagram to prove it, but I won't bother since it's easy to see visually. Each side of the smaller triangle is half the length of the larger triangle's side. The larger triangle is therefore 4 times the size of the smaller one. Edit: Removed my explanation of how to trivialize Towers of Hanoi puzzles, as others may want to work it out for themselves. Edit 2: After looking at other people's answers to 009, it looks like the right answer is going to be D. I maintain that B is also impossible and should also be a correct answer, but D is almost certainly what the game is looking for. And if this is anything like the first Layton game, it won't give credit for alternate answers, so I'd probably change my answer to D. Tax Refund fucked around with this message at 19:50 on Jan 21, 2015 |
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Jan 21, 2015 19:41
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Puzzle 018: I don't know which was the start point and which was the end point. I started from the upper right-hand corner, but I may have gotten start and end mixed up. Oh well. From the upper right-hand corner, head left 1, down 2, right 1, down 1, left 2, up 1, left 2, down 2, right 1, down 2, left 3 and you're at the lower left corner. Incidentally, there's a trick that made this one trivial, but I'll mention it after others have had a chance to solve it. Puzzle 019: There were six people on the train at the start, and one person got off at each stop. There are other, much higher, numbers that could also work, but 6 is the minimum. Puzzle 014: If there were more than four red caps, then everyone (including those wearing a red cap) would have gotten a red balloon. If there were fewer than four red caps, then nobody would have gotten a red balloon. Since only some kids got red balloons, there must have been exactly 4 red caps, and thus 6 blue caps. So the six kids wearing blue caps got red balloons, and the four kids wearing red caps got blue balloons. Puzzle 016: A Puzzle 025: The alligator and the panda don't have a path to the exit. Neither do the tiger or the giraffe. Both the lion and the rabbit have paths to the exit, but only the lion is going to make it out safely. (The rabbit will be inside the lion's belly by that point.) Puzzle 021: Three minutes. In minute 1, A tells it to B. In minute 2, A and B each tell it to one of C and D. And in minute 3, the four who know the message tell it to the four who don't.
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Jan 23, 2015 02:18
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Puzzle 012: In GIMP or MSPaint or whatever, draw thin vertical lines between each pair of white dots along the picture's frame, and one thin horizontal line between the dots on the left and right. Now the picture is divided into ten squares, each with area 1 since the picture has area 10. Call A1 the lower left and E2 the upper right. Now you can see that A1 and B2 are reverse images of each other: mentally swap the clouds from A1 for the sky from B2 and you'll have a completely blue A1 and a completely white B2. Another pair is C2 with D1, and the B1-C1 rectangle pairs up almost perfectly with the D2-E2 rectangle. So between those eight "paired" squares, you can count them as 4 blue squares and 4 white squares. Add the fact that A2 and E1 are completely blue, and you end up with the answer: 6 blue sky and 4 white clouds. Puzzle W02, Moving Tiles: Swap the 0 and 5 in the top row, then ROTATE the 9 at the end of the top row so that it becomes a 6. Now the equation reads 850+476=1326, which is correct. Total of 3 moves, or just 2 if the game counts a swap as a single move. Puzzle W03, Wrong Clocks: At first glance, you might think it will be after 30 hours, when both clocks' minute hands are back on the 6, but that's wrong. After 2 hours, one clock will show 2:02 and the other 1:58. After 12 hours, when it's really midnight, one clock will show 12:12 and the other 11:48. After 24 hours, when it's really noon again, the fast clock will show 12:24 and the slow one 11:36. After 30 hours, both clocks will have their minute hands on the 6, but one will show that it's 6:30 and the other that it's 5:30 -- so 30 hours isn't the correct solution. However, it will need to be a multiple of 30 hours, so our search for a solution can be sped up. After 60 hours (when it's really midnight), the fast clock will show that it's 1:00 and the slow clock will show that it's 11:00. After 120 hours, the fast clock will show that it's 2:00 and the slow clock that it's 10:00. 180 hours will give us 3:00 and 9:00, 240 hours will give us 4:00 and 8:00, and 300 hours will give us 5:00 and 7:00. Now we should check 330 hours, just in case -- but that gives us 5:30 for the fast clock and 6:30 for the slow clock. It's only after 360 hours have passed that both clocks will show the exact same time again, 6:00. (One will "actually" be showing 6:00 AM or 06:00, and the other 6:00 PM or 18:00, but since these are 12-hour clocks that doesn't matter.) The answer is 360 hours. Puzzle W04, Cut and Splice: Puzzle W05, Wall of Dice: Normal dice have a specific layout where the pips on opposite sides always add up to 7, but we can't assume that's the case here. So let's work it out. From the upper right die, we can see that whatever number is opposite 2 can't be 1 or 4, and from the upper left die we see that it can't be 3 or 6 either. So 5 is opposite 2. Now from the upper right die, the number opposite 1 can't be 2 or 4 or 5, and from the lower right die we see that it can't be 3 either. Therefore 1 is opposite 6, which means that 3 is opposite 4 and these must be normal-layout dice. So let's work out the numbers on the bottom left die. Its right side is 5, so its left side is 2. Its top is 4, so its bottom is 3. The front must be either 1 or 6, but which one? Well, let's look at the upper-right die again. If we turn it 90° away from us so that its 4 is on top (matching the lower left die) while its 2 is still on the right, its 1 is now at the back. But the lower left die has its 2 at the left, not the right, so we need to spin it 180° keeping the 4 on top. That brings the 2 over to the left, and it also brings the 1 out to the front. Therefore, the question mark on the lower left die is a 1. Puzzle W06, Paper and Scissors: Draw the circle, then draw a square inside the circle, and its corners will be in the midpoints of the original square. Divide it into quadrants and you'll easily see that in each quadrant, the big square's section is twice the size of the small square's section. (You could also call the area of the big square 4, which would make each side of the small square have length of sqrt(2), and then square that to get 2, but I prefer the quick visualization method.) No matter whether you got it by visualizing it or by doing geometry, the big square is 2 times larger than the small square. Tax Refund fucked around with this message at 05:56 on Jan 24, 2015 |
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Jan 24, 2015 05:46
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Carbon dioxide posted:I'm going to post my slightly different way to get to the solution, because I think it's faster. Yes, this is a better way to do it. I'm usually good at seeing the shortcuts in math-oriented puzzles, but I couldn't find the shortcut this time so I had to work it out the long way. It's good to know what the shortcut was. Speaking of which, that reminds me of a puzzle. (I'm as bad as the Layton characters!) Might as well throw it in as an extra challenge for the thread to have fun with. Puzzle TR01: You are working for a chocolate-making company, in their vending machine department. You have been asked to check on 100 of the company's vending machines, each of which is stocked with 100 chocolate bars. Each bar is supposed to weigh precisely 100 grams, but one vending machine among the 100 has been stocked with bars of chocolate that are precisely one gram too heavy. Your notoriously penny-penching employer wants you to find out which one of these 100 vending machines has been stocked with the one-gram-too-heavy chocolate bars. You have access to a digital scale, accurate to the gram, with a huge platter so you can put as much chocolate as you want on it at once (vending machines won't fit on it, though). Thing is, this digital scale was so expensive the company only bought one of them, and all the other testers need to use it too, so your boss has told you "Use the scale as few times as you can — and if you use it more than 10 times for this test, you're fired." You can push the "vend" button on each machine as many times as you like, though, since that doesn't cost the company money. So how can you fulfill this assignment and keep your job? And as a bonus question, can you find a way to do it in the fewest weighings possible, and prove that to your boss so that he promotes you?
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Jan 24, 2015 14:44
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Accursed Walrus posted:Oh good, I love a good puzzle! Welp. Not having played the games (obviously), I didn't know, but I shouldn't have been surprised that this puzzle showed up in some form in the game. After all, one of my favorite puzzles is the "tell which coin is counterfeit / which weight is off by a fraction, out of 12, in just 3 weighings" puzzle, and that (or a variant) showed up in the first Layton game. I was planning to put my answer in spoiler tags anyway (after giving people about a day to solve the puzzle), as that way anyone else reading through the thread later can still try to solve it. But it's good to have that rule explicitly stated.
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Jan 24, 2015 20:59
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Solution to Puzzle TR01: Nidoking got it right: vend one bar from the first machine, two bars from the second, and so on, up to a hundred bars from the hundredth machine. Then you need only a single weighing to determine which machine contains the overweight chocolate bars. But there's a slightly easier way to figure out which is the faulty machine than subtracting 505000: you can just look at the last two digits of the weight in grams (in other words, take the number modulo 100). If the last two digits are 01, there was only a single overweight bar in the whole stack, so it was from machine #1. If they were 27, machine #27 is at fault. And if the last two digits are 00, then machine #100 was at fault. I have several puzzles that are variants (rather complex variants) on the "who can see which hat?" problem, but I think I'll wait until we've seen a few more puzzles in the game before I post them. Accidentally duplicating one of the game's puzzles has made me a little bit leery.
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Jan 26, 2015 01:47
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Accursed Walrus posted:
The way I solved that one was by this reasoning: First, let's look at the uncut piece. It's 8 squares tall and 10-11 squares wide (10 at the base, 11 at the widest point). I'm guessing that we'll need to cut a more-or-less vertical line, then rotate the part on the right to slot it into place. Now, that little 2-square section on the far right is going to be a problem in most configurations. But, let's see, if I make a cut from the bottom so that there are just 6 squares in the bottom row of the left piece, then that right piece will be able to rotate 90° clockwise to make a bottom row of 8 squares. So that's the bottom row sorted out. Now, if we're rotating 90° clockwise, the next column from the right in the right-hand half, which has 5 squares in it, is going to become the right half of the second row -- which means I need to leave three squares in the left half of the second row. So my second-row cut will move over to the left and then go vertically between the 3rd and 4th columns. There are four of these 5-square columns in the right-hand half, so my cut will go up vertically by 4 rows. And now we're hitting parts where the right-hand half will be only 4 squares high given the cuts I've already drawn, so that means that starting at this vertical position in my cut, I need to go one space to the right before continuing to cut vertically, so that the left-hand side will be 4 squares long. Oh look, I've just hit another border, so the cut is complete. Let me take a look at it, and mentally separate & rotate that right-hand piece... yes, it fits. Okay, puzzle solved. So by finding a good starting point for the solution, I was able to reason out segment-by-segment where each next cut would have to go. Edit: To avoid double-posting, I'll edit my puzzle answers in below. Puzzle 027: It's very hard to make out details at the picture's naturally low resolution, but it looks like the lower right corner of the window is visible to the right of the woman's back, under her cape (or cloak? Nah, it's too short to be a cloak). But that's way out of line with the vertical right-hand edge of the window, which runs along the left side of her hair. I think the unrealistic detail is the out-of-place window corner. Puzzle 041: My intuition says that shape C would fail to form a cube, but I can't manage to prove it. My 3-D rotation skills aren't great. Puzzle 028: Path A is the path that that axle would trace. One large swoop downwards as the point the axle is near heads to the ground, then a small swoop as that point gets rotated around, then a large swoop upwards as the point heads away from the ground. Puzzle 029: This optical illusion makes you want to think that C is the arrow pointing to the sweets, but it's really B. And the fact that B is in a blue circle but points to an arrow with red fletching, while A is in a red circle but points to an arrow with blue fletching, approaches "just plain unfair" levels. Puzzle 036: Tracing the path back from the snake, it connects directly to path A, so A is right out if the bird wants to get out alive. Path C meanders into itself forming loops but never gets anywhere, and path B leads safely to the exit. Puzzle 032: Even with looking at all three hints, this just hits an absolute blind spot in my brain. I can't figure this one out without laying some real wrapping paper in front of me, trying all the possibilities, and seeing which one works. Some people are colorblind, and I'm apparently wrapping-blind or something. Puzzle 034: I almost answered "A and B" because of the image's perspective, but really, A and E are the furthest apart since they're the two endpoints of the straight line of 5 trees. The trick is realizing that the puzzle didn't ask "which two adjacent trees have the greatest distance between them?" Tax Refund fucked around with this message at 05:08 on Jan 28, 2015 |
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Jan 28, 2015 04:26
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Puzzle 037: It's easy to see that the golfer cannot sink his shot in just two putts. At least one of them would have to be 11 meters, and there's no 9-meter putt available. There are multiple ways to sink it in four putts: four 5's, or two 3's and two 7's, for one. So we need to check if there are any three-putt solutions. And there can't be: because his only putting options are odd, three putts will always advance the ball an odd number of meters, but he needs to advance it 20 meters exactly, which is even. (Putting it past the hole with two 11's isn't a workaround for the problem either, as that still leaves him at "-2" (so to speak) meters, which is still an even number.) Therefore, he must use a minimum of four putts to sink the ball. Puzzle 053: There's a way to solve this without trial and error. First pretend we were just wanting to throw out six random candidates via the method suggested. We'll start with the face at the 12 o'clock position and move clockwise, and paint a big black dot on each eliminated face. Then we'll look at the pattern of dots, and we'll notice that it perfectly matches the pattern of women's faces (conveniently placed in pink circles for ease of visual reference), just rotated and maybe flipped. That's because no matter what face you start at, you'll always eliminate faces in the exact same pattern. All we need to do is rotate the pattern of elimination dots, and maybe flip it by choosing to go counterclockwise instead of clockwise from the start. Then whichever face ends up under the "12 o'clock" dot is the face we should start with, and the puzzle's solution. So starting with the 12 o'clock face and moving clockwise, we'll eliminate 6 o'clock first. Then we come back around to 12, which is the next to go. Next we skip over six and eliminate 7, then we skip over twelve and eliminate 2. From there we skip over six and seven and end up eliminating 10, and then the last one eliminated ends up being 8 o'clock. So our elimination pattern is 6, 7, 8, 10, 12, 2: three eliminated circles in a row, then jumping by twos in the direction chosen (clockwise) for three more spots. That pattern matches the women's faces if you put the "6 o'clock" dot on the black-haired women with no glasses in the real 5 o'clock position, and move counterclockwise. Therefore the woman directly across from her — Flora, as it happens — is the one in the 12 o'clock position and is our starting woman. So we need to start with Flora's portrait and then move counter-clockwise. Puzzle 030 UK: Sadly, reading down to the US puzzle kind of spoils the solution. If you fill the "wineglass" part of the sculpture with water and convert it into a fountain, the 2 will have its top half reflected in the water, and will end up looking like a 3. The 1 will look a little ugly, but should still be readable, especially if you take a chisel to its "tail" and make it a simple vertical line. So the town was founded on 13/08, or August 13th. Puzzle 030 US: This one's insultingly easy since they explicitly spell out the "water in the wineglass" part. In the reflected pool, the answer can be seen: after drinking all night, the townspeople all stayed in BED the next day. Puzzle 047: This is a simple plurality (to win, you must get more votes than any other candidate) system, rather than a majority (to win, you must get a number of votes that is strictly greater than exactly half the voters). If write-in candidates were allowed, you could get a situation where everyone voted for themselves except for a single person who voted for his neighbor. Then there would be thirty-eight people with one vote and just one person with two votes, who would beat out all the others. So the answer would be 2 in that case. However, that's not how elections in this town probably work: the puzzle is probably assuming that all the votes will be cast for one of those three candidates and nobody else. Thus, a 3-way tie could be achieved after 39 of the 40 votes had been cast, with each candidate getting 13 votes. In which case, the next vote cast would give someone 14 votes, which is the fewest number of votes a candidate could possibly get and still win. And thus, it's also the answer to the puzzle. Edit: Someone said the answer was 13 votes, because they weren't counting the candidates' own votes. But the candidates are themselves locals of the town, and their votes count too, so the answer is 14 votes, not 13. Puzzle 052: You could start the puzzle like this: Wednesday was a no-hat, cloudy day. If Tuesday had been cloudy, he would have worn his hat on Wed (rule 3), so Tues wasn't cloudy. If Thurs had been cloudy, he would have worn his hat on Thurs (rule 3), so Thurs wasn't cloudy. And so on. However, you'll then start to realize that there's a better way to do it: start with the weather tiles we have available to place, and work backwards. Starting with the weather tiles, we see that there are two rain tiles to place, and only two places we can put them: on Thurs and Sat, the two no-hat days whose weather is unknown. Therefore the four hat days must contain three sunny days and one cloudy day. But we can't make the cloudy day fit in the week (see the reasoning about Tues and Thurs, for example)... unless the last day of LAST week (last Sunday) had been cloudy! In which case Mon was a cloudy day, the second in a row, so the man wore his hat -- and thus the other hat days were all sunny. So from Monday through to Sunday, the weather was: Cloudy, Sunny, Cloudy, Rainy, Sunny, Rainy, Sunny. (Edited once to expand a little on the answer to puzzle 047). Tax Refund fucked around with this message at 09:12 on Feb 1, 2015 |
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Feb 1, 2015 09:09
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Nidoking posted:The voting puzzle asks how many votes a candidate needs to "secure" victory. In other words, the smallest number of votes such that the candidate wins no matter how the other votes are cast. You can't assume that the remaining votes are evenly split between the other two candidates, but must find a value such that regardless of the split of the remaining votes, the chosen candidate will always win. The point in the counting at which the candidate can safely stop paying attention to the reports and start the victory party while the remaining votes are counted. Which makes this a poorly-written puzzle. If you're right about how the puzzle's wording should be read, then your answer is correct. If I'm reading it correctly (what vote distribution yields the smallest possible winning vote count), then my answer is correct. And while your reading of what they mean by "secure" victory is quite plausible, the puzzle could have been made unambiguous by a simple "... no matter how anyone else votes" line being added. But it was left ambiguous, and an ambiguous puzzle is a bad puzzle.
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Feb 1, 2015 17:14
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Accursed Walrus posted:I'll go ahead and settle this, actually. In that case, nidoking's answer is right and mine is wrong. I won't bother editing my post as it's kind of pointless, but if you're counting majority answers on the thread to see if the thread collectively got things right or something like that, then you can count me as giving the same answer as nidoking now.
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Feb 2, 2015 02:53
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Puzzle 051: B is the correct reflection. A has a difference in the eyelids, and C has a difference in the whiskers. Puzzle 038: Take the two fastest horses (1 and 2) to point B. Time elapsed: 2 hours. Ride your fastest horse back. Time elapsed: 3 hours. Take the two slowest horses (4 and 6) to B. Time elapsed: 9 hours. Ride the 2-hour horse back. Time elapsed: 11 hours. Take horses 1 and 2 back to point B and you're done. Total time elapsed: 13 hours. You could also have ridden the 2-hour horse back first, and the fastest horse back the second time, but the total would have still been 13 hours. Puzzle 033: This one's much easier to follow than most puzzles of this type, for some reason. There are 7 fish caught in the net. Puzzle 044: I'm terrible at visualizing knots, but it looks like 2 loops will be caught: the orange and the green ones. The blue should slip out.
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Feb 3, 2015 05:14
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Puzzle 042: Two one-segment cuts, one horizontal cut between A5 and A4, and one horizontal cut between C4 and C3. This will divide the board into an inverted-L tetris piece and the rest of the square. Now slide the inverted L one unit right and one unit down, and you have the square. Puzzle 043: Like 042, make two horizontal one-segment cuts. The first one will be between A4 and A3, and the second one will be between D5 and D4. Without needing to rotate the pieces, the second one can be lifted up, slid one unit right and three units up, and then dropped into place to form the rectangle. Puzzle 050: Let's rephrase the cows' statements to make them easier to read, like so: A: D lies. B: C lies. C: A is truthful. D: E lies. E: B lies. If C truthful, then so is A, and the rest must be lying since there are only two Troomoo cows. But D says that E lies, so that would mean E was truthful... which gives us three Troomoo cows, contrary to what the puzzle states. Therefore C is a liar, and so is A. So D is truthful, E is a liar, and B is truthful. The two Troomoo cows are B and D, and the three liars are A, C and E. Puzzle 039: C is wearing A's red trousers, so A must have kept his red shirt. Since C doesn't completely match and isn't wearing his original clothes either, his shirt must be blue, and B is wearing the white shirt. B can't be wearing the white pants, so they're on A, and B is wearing his original blue pants. A: red shirt, white pants B: white shirt, blue pants C: blue shirt, red pants Puzzle 048: Mr Anderson is the one in the upper left, whose bowler hat just blew away.
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Feb 6, 2015 08:19
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Puzzle W07: 89 + 9 = 98. A = 8, B = 9 Puzzle W08: It looks like you take the difference of the two squares above and put that in the square below, and that would seem to work for all but the upper right-hand section of the puzzle. If that was the case, the question-mark square would be 2. But in fact, the actual answer is to add the digits of the two squares above, and put the digit sum in the square below. Which means the question-mark square should be 16.
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Feb 9, 2015 23:47
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Puzzle 031: Johnny has two more apples than Thomas. Some simple algebra would solve this one, but instead I just started checking all the pairs of integers like 3/5, 4/6, 5/7 and looking at which one fits the other clue... and 5/7 fits it. Thomas has 5 apples and Johnny has 7. Puzzle 046: Put C on the bottom layer to fill it up, then A and B fill up the top layer. No rotations needed from the way the puzzle pieces are presented; I'm sure future puzzles of this type will be harder. D is the piece left over. Puzzle 035: By clue 4, D is not red. By clue 1, blue can't be in C because the red one would have to be D, which can't be. So blue is in either A or B. By clue 3, blue can't be in A since that would leave no way for green to be closer to the power lines, so B is blue. Now D can't be green as that's farther from the power lines than blue (clue 3), and we've ruled out D being red, so D is yellow. Now by clue 1, C is red, and so A is green. Puzzle 054: There are 6 simple permutations: ABC, ACB, BCA, BAC, CBA, CAB. Then there are ABA, ACA, BAB, BCB, CAC, CBC as well, another 6. So there are 12 designs in which all three sections of the flag are dyed. But wait! The rules never say that we have to dye all three sections! What if we left one or two sections undyed? In that case, white becomes another possible color, and we have the following possibilities (designating white as _): A_A, A_B, A_C, AB_, AC_, ABC, ACB = 7 starting with A on the left B_A, B_B, B_C, BA_, BC_, BAC, BCA = 7 starting with B on the left C_A, C_B, C_C, CA_, CB_, CAB, CBA = 7 starting with C on the left _A_, _B_, _C_, _AB, _AC, _BA, _BC, _CA, _CB = 9 starting with white on the left That makes a total of 30 different designs using three dyes plus white. Puzzle 040: First we place the triangle, since the spade tells us its orientation (upwards on the unfolded view) and placement. Now the triangle is pointing at the head of the stick figure, so we place the stick figure one square above it, but looking "upside down" in the unfolded view. Now the spade also gives us the heart placement: the cleft of the heart is pointing towards the left side of the spade. The square that will fold down to be on the left of the spade is the one up and left of it, and that square's low edge will become the edge touching the spade's left. So the heart goes there, also "upside down" in the flat view. Now the club goes in the upper square, and it's "upright" in the flat view so that when it's folded to the opposite side of the cube, its orientation will match the stick figure. Finally, the circle is placed in the only remaining space, just above the spade, and its orientation doesn't matter. Result: http://lpix.org/1941181/Update10-capture_053B_solved.png Puzzle 049: He bought a telephone, which means he needs others to have one as well so he has someone to talk to. Puzzle 055: Take one of the segments and open every single link in it. Take each of those seven opened links and close it in between two of the other link chains. Seven links have been opened and closed, so the total cost is 14 pounds.
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Feb 13, 2015 09:49
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Puzzle 056: X must be 2nd from the top. Top can't be star or moon since they have to be next to each other, and neither can be second. Top can't be diamond since circle must be above it. Therefore, top must be circle. Since moon must be two below diamond, diamond is 3rd and moon is 5th (bottom), leaving star in 4th position. Top: Circle 2nd: the X 3rd: Diamond 4th: Star Bottom: Moon Puzzle 057: Number these squares like a chessboard again, and let's start by thinking about the top left corner. Square B6 can only be filled with fragrance by a rose in B6, A6, or A5: anything else can't reach it. Can we put a rose on A5? Well, its fragrance will reach C5... but then D6 becomes a problem. Because anything that can reach D6 would also reach C5 and become overpowering. So we can rule out A5, because it would block D6. Let's put a big black X on A5. Likewise, A4 and A3 would be a problem, because B6 would be unreachable without putting too much fragrance on A5. So X out A4 and A3 as well, and now we can see the start of a pattern. Any dead end can give us information about where roses CAN'T go: if the rose would leave just a single unreached space, or two unreached spaces, in the dead end, then that rose is impossible. By continuing to put black X marks on where roses CAN'T go, we quickly narrow it down to where they CAN go, as follows: Roses on A6, B2, D6, D3, and D1. 5 roses total. Puzzle 059: I have no clue yet. I have one idea, which I'll spoil.You could maybe fold the ticket along the horizontal line so that the top ends up superimposed on the bottom (though flipped vertically). But I have no idea how that idea gets me closer to the solution. Not yet, at least.
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Feb 18, 2015 06:02
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Puzzle 059: I get it now. You don't fold the top onto the bottom, you just cut out the lighter parts and slide the dark-shaded parts together, which makes the ticket spell out "FOR FOLSENSE". In order to form the letter E with the part above it, the number that was cut out must have bars in bottom-left and bottom positions, but not in bottom-right. The only number that fits the bill is 2 (two).
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Feb 18, 2015 11:03
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Puzzle 061: Took me two hint coins to find it. C is the hotel, for the reasons the second hint points out. Puzzle 062: Start by looking at the dead ends and eliminating squares based on "If a rose was here, it would be impossible to cover the dead end without doubling up on at least one square". Mark those impossible squares with a black X, and suddenly the intersection at C4 looks like a pretty good place to try a first rose. And sure enough, with a rose at C4, the rest all fall into place. Roses at A1, A5, C4, D1, E3, F5, and F6.
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Feb 28, 2015 14:23
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Puzzle W10: Same principle as with all "traverse each line once" puzzles: if there are any odd-numbered points (points with an odd number of lines coming off them), you must start at one of them and finish at the other. If there are more than two odd-numbered points, the walk will be impossible. The cube as shown has four odd-numbered points, so we need to connect two of them, which rules out connection B. And if we made connection A, the starting point (indicated by the arrow) would become an even-numbered point, so that rules out A. Connection C is the answer. Puzzle W11: Another algebra puzzle, solving B+3 = 2R and 2B = 3R. So 4B = 3*2R; 4B = 3*(B+3); B=9. Bill is 9 and Rupert is 6. Puzzle W12: I can't visualize this at all. Update: the hint clued me in. The answer is D. Puzzle W13: The formula includes the term (x-x) in the many factors being multiplied together. The answer is zero, since (x-x) is zero no matter what the value of x, and anything multiplied by zero will be zero. Puzzle W14: Fourth from the top and second from the bottom means there are 5 rows. Third from the left and third from the right means there are 5 columns. There are 25 gems total. Puzzle W15: We'll ignore the fact that a sentence ending in "isn't she?" isn't a statement and technically can't be true or false, and just pretend that statement B was "Mary is older than Lisa." None of the statements obviously, directly contradict each other, so let's look at the six possible ways the true or false statements can be spread out and see which of them "work". If A & B are true, C & D are false. Lisa and Sam are the married couple (since D is false), but Mary (the daughter, by elimination) is older than Lisa (since B is true). Contradiction. If A & C are true, B & D are false. Lisa and Sam are the married couple (since D is false), but since C is true, Sam (the father) is younger than Dan (the son). Contradiction. If A & D are true, B & C are false. Dan is younger than Sam (false C). So Sam must be the father, and Dan the son. Since D is true, Sam's wife can't be Lisa, so it's Mary, and Lisa is the daughter. But Mary is younger than Lisa (false B). Contradiction. If B & C are true, A & D are false. Lisa and Sam are the married couple (since D is false), but since C is true, Sam (the father) is younger than Dan (the son). Contradiction. If B & D are true, A & C are false. Dan is younger than Sam (false C). So Sam must be the father, and Dan the son. Since D is true, Sam's wife can't be Lisa, so it's Mary, and Lisa is the daughter. And Mary is older than Lisa (true B), and Mary is older than Dan (false A). No contradictions in this scenario. If C & D are true, A & B are false. Sam is younger than Dan (true C), so Sam is the son, and Dan is the father. Mary is older than Dan (false A), so she can't be the daughter, she must be the mother. But since B is false, Mary (the mother) is younger than Lisa (the daughter). Contradiction. The only scenario that doesn't lead to a contradiciton is if statements B & D are true. So Sam and Mary are the couple.
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Mar 7, 2015 23:36
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| # ¿ Apr 24, 2024 14:16 |
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Puzzle 089: Draw diagonal lines between the four points of the curved square-ish figure, and "cut" it into four sections. Each section can be slid and/or rotated around to "cap" the lower circle, turning it into a square turned 45°. A little thought will show that the diameter of the circle is a line right through the center of the square, therefore it is equal to the square's sides. A square with sides of 20 meters long will have an area of 400 square meters. Puzzle 068: I had to use three hint coints, but I finally found it. http://lpix.org/1998975/Update13-capture_035B-solved.png Puzzle 072: A shows a patched-up hole in the window of the left building, B shows the hole, and C and D show no hole. A is last, B is just before A, and C and D are the first two shots. To tell C from D, look at the light bulb above (and left of) the right-hand building. It's absent in C, but present in D and the other two shots. C must be first, then the lightbulb was put in in D. Then the window got broken in B, and it was patched in A (but then the neon letter A burned out). The chronological order from first to last is CDBA. Puzzle 117: We don't actually need to know the clue that "no one man is standing directly above his spouse" to solve this one. The husband of the younger sister is wearing no hat, so he's C, and the husband of the elder sister has no moustache, so he's A. So the man the puzzle asks us to find must be B.
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Apr 8, 2015 12:50
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