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140 is somewhat elegant once considered properly: block D must be on the base, so block B must be on top. The other blocks can be fitted into a correct config, but actually doing so isn't needed to get a solution. edit: update on the bottom of page 5! Bacon In A Wok fucked around with this message at 22:50 on Aug 24, 2016 |
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Aug 24, 2016 22:43
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140. 14:00 142. 84 143.A=8 B=7 C =4 145.211 pearls 146.90 degrees
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Aug 25, 2016 06:00
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Yeah, I saw the gimmick in W31 but it clearly should have said something like 'faces on the top of the two dice' instead. #139: The flight arrived in NYC at 17:00 LAX time, and so took 5 hours. It will arrive at LAX at 14:00. (12 + 5 - 3). #140: I thought this was much tougher until I noticed the 'no rotation' restriction. D is right out, as it must have its high point on the outside. Block E also forces D out of the center, since it can't stack with D. And if E is in the center, there's no place for C to go, so E is out. The only block that fits in with the low block in D is A, with D in the 'front'. That means C is out as well, since A will fill the center of the cube, and the answer is B. (In fact, the chunks all appear to be in the exact disposition they'd be in the shape, making it merely an exploded diagram.) #142: You can work this out by looking up how old Diophantus was when he died. He was 84. (Incidentally Diophantine Equations are equations with only integer solutions, like this problem and the next one.) #143: (10A + A) * (10A + A) = 1000B + 100B + 10C + C. =100A^2 + 20A^2 + A^2 = A^2 * 121 = 1100B + 11C. Divide by 11: A^2 * 11 = 100B + C. We require an integer solution (with B <> C, and 0 < A,B,C < 10 - technically C could be 0 but it clearly can't be). My first guess is that the square of A adds to 10, and indeed A = 8 gives a solution. A = 8, B = 7, C = 4, and 88 * 88 = 7744. #145: 2*3*5*7 + 1 = 211. The number desired, of course, was 421 which makes you wonder why the wizard's thought bubble looks so smoky. #146: This is the sort that seems so easy that I'd probably get it wrong at a glance, so I figured it out the hard way: First, find the length of the sides of triangle ABC. Let m be half the length of a side of the cube. Then AB and BC each are equal to /^2 m [ using /^to mean 'square root'.] The length AC can be found by using Cartesian coordinates, and assigning the origin to the point on the base below A. This puts A at (0,0,m) and C at (2m,m,2m). The distance between the two is /^(4m^2 + m^2 + m^2), or /^6*m. Knowing the length of the sides, we can use the Law of Cosines: AC^2 = AB^2 + BC^2 - 2(AB * BC)cos(ABC) 6m^2 = 2m^2 + 2m^2 - 2(2m^2)cos(ABC) 2m^2 = -4m^2*cos(ABC) -1/2 = cos(ABC) Therefore ABC = 120 degrees. The easy/visualization way: If every face is connected by continuing around the cube, there will be six points connected by lines of equal distance, all lying in the same plane. This forms a hexagon, which has interior angles = 120.
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Aug 25, 2016 08:11
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 Pandora's Box (Live Version)  hey so this is cool and good please listen to it!!   3. The time difference has the reverse effect of the flight returning to Los Angeles. If the plane to Los Angeles left New York at noon New York time, you could also say it left New York at 9 a.m. Los Angeles time. Despite the premise, this is still a pretty simple math puzzle, all told.      3. Chunk D forms a part of the front-left side of the large cube. Use that information to start reconstructing the structure in your head. As the pieces come together, the answer should become clear. Normally I kind of have trouble mentally picturing how these fit together but in this case it's pretty easy even for me. I think that they're fixed in this orientation helps a lot, though.     Puzzle 141 Disappearing Act 6  1. You have quite a few balls to get rid of, so you'll have to do a lot of jumping. Just make sure you aren't choosing your moves at random, or you'll end up getting stuck before you finish. Plan out your jumps so that you never leave any balls out away from the group where you won't be able to reach them later. 2. As you reduce the number of balls on the board, those remaining take on a configuration that you've previously solved. If you made it this far, this puzzle is in the bag for you. 3. Sorry, there are no more hints to be had for this puzzle. You may have heard that proper peg solitaire players don't consider these puzzles complete unless the remaining ball ends up in the center space. One step at a time, though. Let's finish this puzzle first. Okay, so this is the closest thing there is to a proper standard Peg Solitaire puzzle at long last. Still don't have to end in the middle but the easiest solution makes that available anyway so why wouldn't you?  End  3. The lowest common denominator of 2, 6, 7, and 12 seems to be the same as the length of the mathematician's life expressed in years. As written, this puzzle was very convoluted and confusing but when you cut the chaff it becomes easier to understand.    3. As detailed in Hint Two, if you substitute every digit for A and work your way up, you'll find the answer in the last half of your search. This one is honestly kinda hard to figure out simply, but very easy to figure out quickly. Just hammer combinations in a calculator until the number you square results in a BBCC answer.  Puzzle 144 Super Pancakes  1. The method for solving pancake puzzle has been described in many other hints. Move the smallest pancake over to the middle plate, then move the next-smallest cake to the right plate. Then stack the smallest cake on top of the one right. Next, take a new, slightly larger pancake from the stack on the left and move it onto the open middle plate. Read more on what to do next in Hint Two. 2. From where Hint One left off, put the smallest pancake on the pancake sitting on the middle plate. Then move the pancake on the right plate over to the stack on the left. Bring the smallest pancake you have over to sit on top of the stack of pancakes on the left plate. You can solve this puzzle by repeating the process outlined in Hints One and Two while introducing bigger and bigger pancakes, but it's not very efficient. 3. Aim to limit the number of moves you make to keep yourself from getting confused along the way. Being just another Tower of Hanoi puzzle, this isn't particularly difficult or anything. But the thing about Towers of Hanoi is that they get really long, really fast. Based on optimal solutions, this one is by far the longest puzzle yet. ...Based on (my) average solutions, it pales in comparison to most sliding puzzles.     3. The number the magician was after is greater than 400, but the number of pearls the adventurer brought back was actually in the range of 200 to 300 pearls. Remember, you're out to find the number of pearls the adventurer decided to bring back. Adventurer failed his quest because he failed his maths classes.  It's really not hard to get 421 as the intended solution! 211 is obviously wrong, because that gives 3 left over when divided by four. 212 is so obviously a multiple of 4 that it's astounding!   3. Think about the regular geometric figures you can create with points A, B and C. You could create a regular hexagon, couldn't you? And the internal angles of regular hexagons are always 120 degrees. Mentioned it before, but the lack of any numbers make this easier than it would be otherwise, since it helps remove the risk of overcomplication. This is a real risk of happening with all math puzzles, as has been witnessed countless times in the past.  Also wow I just noticed but Hint 3 outright tells you the answer. Not even in a coy wink-wink, nudge-nudge way. It just spells it straight out. You'd think I'd have noticed it before but I hadn't looked at it before so... Puzzle 147 Sliding Labyrinth  1. Just moving pieces around randomly probably won't help much. Think about the position of pieces on the grid as dictated by where their tunnel segments are placed. 2. The piece connected to START! is the straight horizontal tunnel. Leave the bottom right square on the map open. 3. The pool is below the start section. The cross-shaped tunnel goes in the center of the map. Honestly, for an intended to be hard sliding puzzle this one is... really easy. It should not at all be worth 80 Picarats because it's far easier than a lot of earlier ones!  So, next up on the list is the first actually real postgame set of puzzles.  And unlike before, all 3 of these CAN be done in-thread!   1. The product has four digits, the first of which is one.   1. Try treating each digit of a number as a separate entity. 2. Try to find a rule that uses the digits from one number to create the next number in the cycle.   1. This board is eight spaces by eight spaces, just like a regular chessboard. That means this puzzle is just like the original knight's tour. There are no hints to be had here! Just work it out on your own. Yeah, so I did all the main-game versions of this without thread input and now you get to handle the hardest one on your own! ...In part because it's doable like that and in part because I'm sure someone probably wants to anyway.  For the sake of having more puzzles, let's pretend that those 3 have been solved now and that brings our total up to 150. That lets us access the last set of 3.  Curious Village made these worth relatively low amounts for when they came.  Thankfully, that's not the case here.  1. This is a hard puzzle, so maybe this hint will help. The first thing you need to do is create an answer key based on the three scored tests. Since Mary got the highest score, try creating a key that compares her answers to everyone else's. 2. The three graded tests agree on four questions. However, you know that not all four of these are correct because Lisa only got a score of 30. You also know that not all of these "agreed" answers are incorrect because Mary had a score of 70.  So this set of three aren't all worth 99 (even though that'd be sensible for the placement!) but whatever. It doesn't matter; by now we have more than enough Picarats for everything they're used for anyway.  1. Mock up an equation with the number of people and the number of matches played as variables. That would be a good place to start. 2. In order to create the equation for the number of hands, you would need to multiply the number of players by the number of players minus one, then divide this product by two. If there were three people, there would have been three hands. If there were four people, there would have been six hands. You can assume from the conditions given that if no one left, the minimum number of hands would still have to be at least 60. Of course, 153 is a sliding puzzle. The last puzzle of (almost) every set here has been one so it makes sense, but also Curious Village's final puzzle was a sliding puzzle too. I can't really complain about it being consistent I guess.
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Aug 25, 2016 23:12
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151: Dan's answers to questions 6 and 9 are right - he had the same answers as Lisa on all other 8 questions, and scored 20 points more than her. So Mary's score of 70 is right answers on all but 6, 9, and one of 1, 3, 7 or 8 (doesn't matter which - John's answers to these four questions match Mary's). John's answers match Mary's for seven questions (where John got five correct, two incorrect) and differ for three questions (where John got one correct, two incorrect). John's score is 60. ... also, what kind of school has neckbeard Dan and babyface Lisa in the same class, anyway? 152: Total number of tournament rounds for an n-player round-robin is of the form 1+2+...+(n-1). 59 rounds played with one player skipping out on some rounds suggests a starting pool of 12 players, 66 rounds in a complete tournament (11 games per player) so Monty played 4 games before leaving.
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Aug 26, 2016 01:02
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Huh. I solved 140 about like so: each block has one block on the uppermost layer, and you can't spin them, so the chief block must have at least 5 blocks one tier down so the other 4 can fill in the holes. 149: Each number is the sum of the squares of the 152: 12 people played (12 * 11)/2 = 66 hands total ideally, meaning Monty missed 7 of his possible hands, leaving him to only play 4.
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Aug 26, 2016 01:07
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Professor Layton's Theme (Live Vers.) hey its awesome and its even relevant and appropriate here!!   2. The last digit of the product is six. 3. The single-digit multiplier is three. I dunno if I'd call this puzzle exactly hard or anything. Just... kind of weird and eh. It's a math thing, but it's just odd about it I guess.    3. Try squaring the individual digits of each number first. Conversely, this one seems weirdly convoluted until you see the 20 -> 4 -> 16 part and realise what you're meant to do.  Puzzle 150 The Knight's Tour 4 2. The knights tour usually involves returning the knight to the starting space at the end of the tour. If you're trying to accomplish this, then choosing the correct starting space is very important. Why not use the Memo function to help you? 3. There are many avid chess researchers in Europe and the U.S. who write detailed papers with elegant solutions to chess puzzles. Looking through resources is on more way of enjoying a chess puzzle. In theory, this could be a difficult puzzle for sure. It's fairly lengthy regardless but it's just so well known that the solution isn't a big mystery.  End  3. You can make multiple answer keys that will all yield the same results. As long as you find one of these keys, you can solve this puzzle. There's a couple possible ways of getting the solution. They all give the same answer in the end, though.    3. There were 12 people at the tournament. Now think through! Simple algebraic equation for maximum number of games makes it really easy to figure out the answer here.  Puzzle 153 Ultimate Escape / The Diabolical Box ...The US version of this puzzle is identical but its name is The Diabolical Box. Way better, in my opinion.  1. The difficulty of this puzzle is truly diabolical! You just have to find ways to fit the blocks together to open up a little bit of space at a time. 2. You can't move the gem at all first. You should focus initially on making room for the gem to move. Here's one small hint: move the yellow vertical bar to the top half of the puzzle. 3. Here's your last hint. You never have to move the cross-shaped piece any farther right than the center of the puzzle. In other words, the gem has to move clockwise through the puzzle. Stick with it! Yeah, uh, this is definitely a very difficult sliding puzzle. The video linked above is the optimal solution but it's not my original run for this LP. Oh no. That's this one.  Regardless of how bad my memory can be at times, with this one done as well that's all 153* puzzles in the base game done.  Unfortunately, we've also done all the DLC puzzles already too so, uh, that suggestion is nice but wasted!  Also Layton and Luke now take this moment to break the fourth wall.  It's clear that you possess an aptitude and passion for puzzle solving that is equal to my own. This is your reward for solving all 153* puzzles. It's not really worth it, but there's some good stuff in there. Indeed there has, Luke. That's a gift from us to our friend, the player. That's you! You heard the professor. Have fun with those!Yeah, so there's not much left for us to see here unfortunately. We've done every puzzle* and so we might as well check out what our 5500 Picarats give us.  To do that, we head into the Bonuses screen again but this time we are going to check out that Top Secret section.  Profiles and Art'll be in a bonus update later, since there's a lot in there. Music is hopefully self-explanatory as is Movies. Voice is just a random selection of voiced lines from throughout the game; there's no pattern or rhyme to what was chosen or anything. But the Hidden Door is something we always have access to and it's... different. It's also something in Curious Village we never looked at, because we couldn't!  You see, you need that passcode from the top screen here to access the content in Curious Village. Now, it also has a password for this game so we should go get that. Oh and I'm sorry to disappoint you, but that password is unique to each DS so you can't just copy this one and have it work. Annoying, really.  But let's put that one to use. That means we need to jump back to a Curious Village savefile that has beaten the main game at the very least.  The Hidden Door in Curious Village is just a password prompt. It's interesting how even with this games inception they were intending to make a second one. In an alternative universe, Layton never took off so this content was forever locked off because Pandora's Box was never made.  Once we put that password in, we get access to the password to use in Diabolical Box (which is, again, unique to each DS). We also get a blast from the past with those old portraits. Indeed. From here, you can take a peek into the process of how we went from the page to the screen. It's only a fraction of the material created for the game, but we do so hope you enjoy it! Yeah, I'll just put this here since it's actually kind of interesting and there isn't much of it. Feel free to peruse at your own leisure.          We look forward to seeing you for the next adventure in the Professor Layton series!It's weird how that was written on the presumption you've only ever played Curious Village but you can't access this without beating Pandora's Box so...  Speaking of which, we'll head back to that now and take the password from the Concept Art gallery to input here.  We've got a special treat for you here that we hope you enjoy! Have fun! Puzzle ??? A Curious Present So, if we include this, the base game of Diabolical Box has 154 puzzles! This is your reward for finishing the first two Layton games...  1. This puzzle is a variation on the classic 15-piece slide puzzle you may have seen at your local toy shop. It's not a particularly difficult version of the puzzle, so relax, take your time with it... and have fun! And it's a sliding puzzle. At least it's a fairly easy one, all things considered. Random note, but despite the puzzle being documented reasonably well all over the internet at this point... the Hint is, for some reason, written down nowhere that I can find. Weird.  End This puzzle isn't included in your Puzzle Index, but you can come back here to solve it whenever you wish. Have fun, and do enjoy the rest of your time playing Professor Layton and Pandora's Box.Y-yes, the rest of my time playing Pandora's Box. Because there's stuff left after this. ...Well there is that other thing in the Hidden Door I guess. But that explicitly requires something we don't have yet. That raises a question of its own, though. Where can we get a "Ticket to Time Travel"? I suppose the only way is to go on a strange journey through time itself to potentially find a lost future... 
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Aug 27, 2016 18:55
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It's nice to see a game hand off to the sequel, and rare but neat to see it take a handoff back in return.
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Aug 30, 2016 05:38
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When you do go on layton and luke's excellent adventure, I hope you link it in this thread.
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Sep 2, 2016 15:12
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