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W10: The blue one to the left of his eye. W11: This is algebra! Effectively, cw+ch = 30, and 4*cw + 2*ch = 100. By doubling the first equation, we can subtract it from the second one: 2*(cw+ch) = 2*30 2*cw+2*ch = 60 4*cw + 2*ch - (2*cw+2*ch) = 100 - 60 2*cw = 40 cw = 20 Now we know there's 20 head of cattle on the farm, so there must be 10 chickens. W12: Nope, not doin' it. W13: OK, to avoid confusion, let's refer to each alien by the second letter of their names. So, A, E, I, and U. To recap the constraints: A is farther right than I. U is farther left than A. E is farther right than I. Then, we have: The one two to the right of U isn't E. So we have a few possibilities. If U is on the far left, E must be on the far right. Because, E is farther right than I, so E can't be up next to U, and E isn't two spaces away from U either. That leaves: U ? ? E, which has to be U I A E. So A is two spaces away from U. If U is on the second left, E must be second on the right, and I must be on the left. So A is two spaces away from U. Either way, it's A, the Panporo. W14: The one on the top right. W15: Let's say there's only one sapphire. There's more than twice as many emeralds as sapphires, so there's minimum 3 emeralds. And there's more than twice as many diamonds as emeralds, so there's minimum 7 diamonds. That's at least 11 gems. W16: The only numbers of white, black, and red gems that work are 1, 2, and 4 respectively. So there must be 3 blue gems. W17: OK, so in the third year, there's 300 students. In the fourth year, they admit 110 students but 100 graduate, so they have 310. In the fifth, admit 120, grad 100. So 330. In sixth year, admit 130, grad 100. So 360. In seventh year, admit 140, grad 110. So, 390. In eight year, admit 150, grad 120. So, 420 (blaze it). W18: D. It's mirrored. W21: Math again! A deck has 26 black cards. The first pile has 4 red cards, so 13 black cards. The second pile has 6 red cards, so 12 black cards. 13 + 12 = 25. The third pile must have 26 - 25 = 1 black card. W22: 4 minutes: 4 candles 5 minutes: 1 candle 8 minutes: 5 candles 10 minutes: 2 candles 12 minutes: 6 candles 15 minutes: 3 candles 16 minutes: 7 candles 20 minutes: 4 candles, then 8 candles 24 minutes: all 10 candles 25 minutes: 7 candles W23: D. It looks like a gentleman's top hat! W25:  W26: I got nothin'. W27NA: A-2-3. W27EU: Entrance E. W29: D is a mirror image of C! It's a reflection in a window. W30: C and E. W31: You only weigh once! Put [C T] and [C S] on one side, and [T S] and [D D] on the other. The T and S cancel out, and you're left with C C vs D D. W32: 6. It's divisible by 1, 2, and 3, and 1+2+3 = 6. W33: Balls.  OK, so there are 36 black balls around the outside. Ignoring the 4 in the corners, that's 32 left, so 8 to a side since it's a square. So there's an 8x8 square of white balls inside the black balls. So, 64 white balls.
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Jan 28, 2020 04:21
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| # ? May 4, 2024 00:00 |
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W10 - The proper response is, "Thanks, you shouldn't have!" (upper right eye, striped green oval one) W11 - 4(cows) + 2(chickens) = 100, cows + chickens = 30. Solving this system gives 20 cows, 10 chickens. W12 - Put the 1 to the left of the two, then the + between there and the green 6. Place the 6 on top of the green 6, then the = to the right of it, then the 12. This makes a valid equation... when viewed upside down (21 = 9 +12) W13 - Both Pan and Pen are to the right of Pin, and Pun is to the left of Pin, so from the left it must be Pun, then Pin, then the other two. That means if the alien in question isn't Penporo, it can only be A Panporo. W14 - They seem to like these, don't they? Looks like the purple carpet in the upper right corner to me. W15 - Assuming at least one of each jewel, then there's at least one sapphire, and d >= 2e + 1, e >= 2s +1. If s is 1, then e is 3 and d is 7. 11 jewels in all. W16 - The white:black:gem ratio is 1:2:4, but since their sum can't be more than 10, that's the actual number. That leaves 3 Blue ones. (W = 1, Bk = 2, R = 4, Bl = 3). W17 - After year 4, the number of students in year n in excess of 100, for each class is 10(n-3), 10(n-4), and 10(n-5). We need the sum of that to be more than 100. That gives is 10(3n - 12) > 100, which reduces to n > 22/3, or Year 8 since it must be an integer solution. W20 - D. The eyes and the tongue are the only differences. W21 - There are (17-4 =) 13 black cards in pile 1, and (18-6 =) 12 black cards in pile 2, giving us 25 in those two. Since there are only 26 black cards in the deck, that leaves just 1 in the pile on the right. W22 - Through 15', each time the lighter acts, the extinguisher acts afterward and leaves only one additional candle, so there are 3 lit at 15'. At 16', the lighter acts and 7 are lit. At 20', the extinguisher puts out 3, then the lighter lights 4, so 8 are lit. At 24', all 10 are lit. Then at 25', the extinguisher puts out 3, so 10 are lit. W23 - D, they are seeing his top hat, that he never takes off unless the love of his life has to disappear due to a W25 - Take out the two middle segments in the left and right 'ladders'. A and C will switch tracks to the other side but return, and B and D will go around the side then end up on the inside at the bottom. W26 - Cut the piece in half horizontally, then cut between the two middle hats. Then line up the piece on the right over the two left hats on the bottom, and the piece on the left over the one farthest right. W27$ - Each card has two more hearts on it at the corners, so it is A,2,3 to give 3+4+5=12. W28€ - You can only go right, then left by entering at A, E or F. Of those, only E has a room in the opposite corridor with an exit to the right. W29 - D, maybe? If this is supposed to be a corner, maybe that line is meant to show where the corner is and D should be rotated around it if it were real. W30 - C and E can be doubled to form a shape with an I in the negative space. W31 - 1. Put the two circle boxes on one side and the other two on the other side. W32 - The smallest perfect number? We can use the Euclid-Euler Theorem to get 6. Thankfully they didn't ask us to find the largest. W33 - A square with side length n has 2(n-2)+2n balls for its perimeter. We thus have 36 = 2(n-2) + 2n => n = 10. The white square has side length of 8, or 64 white balls.
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Jan 28, 2020 20:46
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Jan 30, 2020 08:34
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Wow! We're just about done! 154: 3. If there's 23 red cards in this stack, there's no black cards, so there's 26 in the other one. Let's say we move one black card over. Then there's 22 red cards in the short stack, and 25 black cards in the tall stack. And so on... 157:  The other two don't work. If you try to place piece 2, you won't find a good spot for it that lets the other two pieces fit. 158: 10 minutes. This is because, the stations are actually arranged, not like this: A B C D ...but like this: A D B C It's 15 minutes from A to B, passing through D. 5 minutes from B to C. Then 10 minutes from C to D, passing through B. Now, we know the B-C leg of the trip took 5 minutes, so we know that D must be 5 minutes away from B. And since A to B is 15 minutes, A is 10 minutes away from D. Yeah. The last stop isn't at the end of the line! 160: Nope. I have never been good at this particular puzzle. 163: 18:49 x 3 = 56:27 164: There's a clever trick you can use to solve this. The numbers in such a square, assuming they touch on the calendar page must be S (the smallest number), S+1 (to its right), S+7 (below it), and S+8 (below it and to the right). Or, to put it more simply: 4S +16 = 88 4S = 72 S = 18 The number is 18. I take it Puzzle 165 is the traditional super-mega-hard sliding puzzle?
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Jan 30, 2020 09:15
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160. 16, if you make a star in the middle.
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Jan 31, 2020 03:38
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#154 If there are x red cards in the left stack, there are 26-x red cards in the right stack, and 29-(26-x) black cards. That simplifies to x+3 black cards, and the difference is 3. #157 All of them fit, just not all at the same scale. A: 2 in the top, 3 on the right, 1 on the left; B: 2 on the right, 3 flipped on the lower left, 1 on the upper left; C: e: Realized my solution for C won't work as stated; you have to put piece 1 in the notch, and I don't think there's any way to get it to work, so I guess everything but C. Although looking at the 'stamp' handles it's still unclear how much flipping is allowed, so not all those answers may be allowable. #158 20 minutes. B is between C and D. C->D is 10 and B->C is 5, so B->D is 5. A->B + B->D = 15 + 5 = 20. #160 16. Each cut/string can cross the previous lines, adding n new segments. 1 string => 2, 2 strings => 4, 3 strings => 7, 4 strings => 11, 5 strings => 16. #163 A must be 1 to not exceed 60, and D must be 9 to get 7 seconds in the rhs. We also know C,E, and G can't be more than 5, which means B and F are 6 or 8. If B were 6, then with F=8, there can't be any carryover minutes , but with CE at minimum 29, that can't be the case (there cannot be 10 minutes carried over either). So B is 8, F=6, and E=5. That requires 2 carryover minutes, so C is 4 and G is 2. 18:49 x 3 = 56:27 Nice to see an equation that doesn't rely on trickery for once! #164 18 (it seems implied that this is a hypothetical circling, so it would seem that even though the 18 square is visible, it's a valid answer). A calendar square circles n, n+1, n+7, and n+8, which has the sum 4n + 16. If 4n + 16 = 88, n = 18. Kangra fucked around with this message at 18:07 on Jan 31, 2020 |
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Jan 31, 2020 17:37
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Quackles posted:I take it Puzzle 165 is the traditional super-mega-hard sliding puzzle? Fortunately, no! 167 and especially 168 on the other hand...  
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Feb 3, 2020 01:45
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quote:Thank you very much for reading. I hope that sometime, somewhere, we'll meet again. <3 That was a magnificent Let's Play. Thank you for your hard work.
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Feb 3, 2020 09:41
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Thanks so much for showing us this game! It was a lot of fun.
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Feb 9, 2020 21:54
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Back when I was 9, I had just discovered the 'trilogy' idea and while I wanted to see a solo Luke adventure, this was a nice thing to end on. I only later became aware of the prequel trilogy, and while that's fine, has it's charms and merits, I still like Lost/Unwound Future as a nice note to end on (nothing against Alfendi or Katrielle, but those are more spin-offs than sequels.)
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Feb 9, 2020 22:07
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| # ? May 4, 2024 00:00 |
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Luke and Flora sequel please.
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Feb 28, 2020 03:09
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