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I was sort of surprised at the answer to 31, Pass the Apples. The problem is at best ambiguously worded:quote:Johhny and Thomas are each carrying some apples. If Johnny gave Thomas one apple, the two men would each have the same number of apples. Conversely, if Thomas gave Johnny two apples, Johnny would have three times the number that Thomas has. While most of the time people clarify by adding 'now', that is the number that Thomas is currently carrying. Which leads to a solution of 4 for Johnny and 2 for Thomas. Maybe it's intentional, but I would have expected them to try to trick you with strict grammar than rely on a looser, vague inference of what they intended. (Note that in the solution they explicitly leave out the verb, which is what you'd normally do in a sentence like this).
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Apr 8, 2016 17:17
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| # ¿ May 17, 2024 01:49 |
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56:With X in second position, neither moon nor star can be alone at the top. The diamond couldn't be above the circle, so it's not there either, leaving only the circle in Position 1. The only place for the moon to go that's two below anything is the very bottom, putting the diamond in Position 3 and the star in Position 4. W06:The diameter of the inscribed circle is equal to the length of the original square; when we cut to form the second square we've effectively trimmed off the corners of the original on the lines connecting the midpoint of the sides. This cuts each one of the four quadrants of the square in half, which means the new square is half the size of the first one. W07:AB+B = BA where A & B are decimal digits is equivalent to (10A + B) + B = (10B + A) where A & B are integers in [0,9]. We also require A =/= B. The equation simplifies to 9A = 8B, and it's easy to see that the solution is A = 8 and B = 9. W08: For any two numbers x & y, the rule is | x - y | modulo 495, so the answer is 2. Or fine, take the sum of the digit sums and the answer is 16.
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Apr 9, 2016 05:13
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61: The hotel is the one that says HOTEL on it (C). W16: I saw this almost right away, but it's hard to put in words clearly: Die 1 (2-3 showing) and Die 4 (3-2) are just flipped from each other, so that the face connecting to Die 4 is the same as the one connecting 1 & 2. That's also the face on the left of Die 2 (4-5), and the face on Die 2's right is what we need to connect to. Since C is Die 2 flipped over, it will match Die 4 and Die 2 and thus is correct. Another way of looking at it: When solved, Dice 3 & 4 are what you get if you spin Dice 1 & 2 around and over together. Kangra fucked around with this message at 02:43 on May 1, 2016 |
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May 1, 2016 02:39
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These are still not that hard. Do they expect you to freehand the geometric mean of the radii in the plaza puzzle? #129: C is the only point with an odd number of exits, and since there is the weird loop between D & F, it's the only possible place to answer. #76: Logic does not dictate that, but since they're using flawed phrasing the answer is probably the tower itself. #119: Three , unless the bags are allowed to be opened and the coins removed for weighing, in which case it's one.
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May 21, 2016 23:02
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#110: Assuming that the exit is the lower right, and assigning directions based on North=up, 1 East, 2 South, 2 East, 1 South, 1 East. #111: The bobbies who claimed an odd number of turns are lying or violated the U-turn rule. #092:A pentomino can be located in the lower right section that covers gold, silver, and two coppers for a value of 10. #064:There's a maximum at either extreme - either 2 or 100 stones give a result of 51/101. Since 2 is just as good and slightly easier, go with 2. #071: 20 = 2*G' (G' is girl's previous age), girl's current age = boy's previous age : G'+x = 20 - x. From the first equation, G' = 10, and that gives 5 for x, yielding 15 for the girl's age #069: This one is just silly; there's no earthly reason for someone to make copies like that. It can be done in three passes by folding in to get A-D, and then B-C and E-F on their own page. #096: A has too long a tail; C has backwards hands; D has too few fingers on the hands. The answer is B. #074:Hit the corks with a hammer to smash open the two containers and throw away the garlic. Otherwise the next poor person to enter the room will have to deal with it, and there's no way we're getting the garlic out through that mass of tubing. A much more entertaining set this time around! Kangra fucked around with this message at 06:41 on May 24, 2016 |
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May 24, 2016 06:25
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87: Using R for row and C for column: With the initial positioning, only a spade can be placed in C3R3. That leaves only C4R1 for a spade in Row 1, and the diagonal forces a club to C1R4, meaning a spade is in C1R2. Also, C2R1 is a heart. That's Row 1 and Column 1 filled. From there, it's relatively easy to fill Column 2 (HCDS) and then Column 4 (SDCH), and finally Column 3. Completed: DHCS SCHD HDSC CSDH 66: The answer is 2714. Interestingly enough, there's a symmetry between the two (the blue value for a given red is the same as the value to set red to get the blue equal to the original red value). I haven't given it enough thought to see if that's a necessary outcome. 78:The bottom center room currently has no way to exit, so one of those doors must switch. If the red door switches, then the maze entrance has no exit, so that can't be it. If the purple door switches, the lower left room has no exiting door. Therefore, it must be the cyan door in that room (leading to the center), and that will work. This one gave me fits until I realized you don't have to visit every room only once. 81: Clearly this safe combo was set by Lgnikoo Gssal.
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Jul 29, 2016 18:36
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#82: Connecting top part to bottom: 1 to 4, 3 to 1. #77: e: nm, that approach made no sense and was almost certainly wrong. #121: Starting from the upper left: C1R1, C2R2, C3R4, C4R3 covers all rows and columns, plus the diagonals. #097: Label the blocks from the upper left as A, B, C, etc. Then A, C, and G are fixed; the others can move. Since Block I has two stars in every direction, it can only be placed in the center or left of it. Also, Block F must go in the top middle, since that's the only way to finish the top row. This means Block I is on the left, and from there we can get to to solution: A F C I B D G E H #125: I find it harder to notate this solution: Using rough compass directions, head E, N twice, SW, NW, SW, NNE, NE, SE, N twice, W twice, NE twice, S, SE twice, then N and out. Kangra fucked around with this message at 09:54 on Aug 3, 2016 |
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Aug 2, 2016 19:54
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Fans of this update would enjoy the Tilt mazes of Robert Abbott and Andrea Gilbert. #126 - N, E, S, W, N, E, N #127 - N, W, N, E, S, E, S, E should put us in the bottom right corner. Then N, W, S, E, S, E, N to exit. #128 - Easier than the last one. N, E, N, W, S, W, S, E, S, E, N, W, N. #100 There must be something about the mechanics that a screenshot isn't conveying. Reading the hints only makes it more confusing.
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Aug 8, 2016 06:08
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I even noticed the 'go back to start' trick and assumed since it so easily solves the first and second one that it couldn't possibly be allowed. #130 : The most unusual thing is VAMPIRES SHOULD NOT CAST REFLECTIONS! By the way WTF ACTUAL VAMPIRES?! Also I think the picture's viewpoint is upside-down, based on the dresses on the sides. #131: The post on the right; the path to the 'outside' of the figure crosses an odd number of lines. #88: 6 jumps, 12 if he has to follow the purple path. Photo #9: Bent hand railing, flipped track switch, extra rail? piece of bar? radiator element? at back right Kangra fucked around with this message at 05:12 on Aug 16, 2016 |
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Aug 15, 2016 03:01
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What Folsense must have looked like. #29: The turn direction doesn't really* matter, since any given route can be reversed. If we number the nodes (intersections) from top to bottom, left to right, we have point C at Node 4, D at 6, and E at 10 (A and B are not at nodes, and so can be ignored). The starting point must be E, since it has an odd number of exits. Using the numerical notation, one path (making right turns) is 10 -> 9 -> 10 -> 8 -> 6 -> 7 -> 9 -> 8 -> 5 -> 3 -> 6 -> 5 -> 1 -> 2 -> 4 -> 3 -> 1 -> 2 -> 7. *The rules don't make it clear whether you can circle literally any point; if you could, of instance circle only E but not my node 7, then it matters.
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Aug 22, 2016 03:42
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| # ¿ May 17, 2024 01:49 |
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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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