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#47 : Abel can only guess his color correctly if he sees two red hats. Therefore he either sees two white, or one red and one white. Benjamin does not see two red either, and is similarly unable to guess his color. But if he were to see a red hat on Paula, he would know for sure that Abel is seeing a white hat on himself, since Abel cannot be seeing two red hats. Therefore he must be seeing a white hat on Paula. Paula is wearing white, and we don't know what the others are wearing.
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Nov 21, 2020 13:58
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#57 If Abou eats oats, then the cow and horse must both eat oats or both eat hay. If the former, then #2 is false. If the latter, then #3 is false. Therefore, Abou cannot eat oats. The question is unsolvable if there is an option other than oats or hay to eat, so Abou must always eat hay. The actual possibilities for Abou, horse, cow are : hoo, hhh, hho It's also unclear if it's meant to be 'who eats the same food always' or 'which animals eat from the same bag', except that only the first is answerable.
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Nov 30, 2020 07:08
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#62- With the amounts after each pour indicated as (5-cup,7-cup,12-cup): 12->7 (0,7,5) 7->5 (5,2,5) 5->12(0,2,10) 7->5 (2,0,10) 12->7 (2,7,3) 7->5 (5,4,3) 5->12(0,4,8) 7->5(4,0,8) 12->7(4,7,1) 7->5(5,6,1) 5->12(0,6,6)
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Dec 1, 2020 22:00
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#65a 4 steps. Fill 5, pour into 7, repeat. There will be three drops remaining in the 5-drop vial. #65b 6 steps. Fill 7, pour into 5. Pour out the 5 and move the two drops from the 7 into the 5. Fill the 7, pour into 5 and there are 4 in the 7-drop vial. And then fail due to surface tension making it impossible to do any of that.
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Dec 4, 2020 04:53
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Puzzle 67: Using one of the vats, fill the 5-pint container 10 times and dump it out each time. Do this 4 times with the 4-pint container and the same vat. There are now 14 pints remaining in the vat. Fill the 4-pint, pour it into the 5-pint. Do this again, and when the 5-pint is filled, there will be 3 pints in the 4-container and 6 in the vat. Fill the 4-pint container once more, and pour 2 pints to fill the 5-pint, leaving 2 in the 4-pint container. Dump out the 5, and fill it with the last 2 from the vat. So why two vats?
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Dec 6, 2020 08:36
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I was curious whether the vat problem was meant to imply you couldn't drain the vats, but it didn't say anything about it. In 81, there are 30 fish because there were 30 originally expected, but only 28 showed up. That complicates things slightly. You can still use C(30,2), considering the 'pond' as taking two turns at getting a fish. So it's still 1/435 The reasoning below is wrong because the cases where the pond has one fish left are more likely. In essence, they should just be left in and the numbers still work out. e: Actually, it doesn't change things, which was my first thought, but I was half-awake and second-guessed myself. Kangra fucked around with this message at 19:13 on Dec 7, 2020 |
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Dec 7, 2020 17:38
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Puzzle 83 If B is telling the truth, then F is not guilty, implying E is not guilty, implying A is not guilty, implying C is guilty. D's statement is false if B's is true, and so is C's, so that case is consistent. If B is lying, the F is guilty, implying E is guilty, implying ... that's already three guilty, so we conclude that B is telling the truth. Thus C and D are guilty.
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Dec 9, 2020 00:01
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Puzzle 83 No one can both be guilty and have their one truthful statement blame another person or deny their own guilt. That means none of A's statements can be the single truth of a guilty person. So A cannot be guilty. B-1 and B-3 cannot be B's single truth if guilty. B-2 could be, which would make B guilty. e: Just re-read the problem statement, and C is known to love the pie, so C cannot be guilty. We can infer that C probably would not steal the pie if he doesn't like it, however, so even if it's not logically forced, I imagine B is more likely to be guilty here. There's also the neat logic that if B is guilty, all of C's statements are likely true, and all of A's statements are likely false, which would be more consistent with the other citizens. (If C is guilty, B has a mix of one true, one false, and one probably-true.) Kangra fucked around with this message at 19:01 on Dec 10, 2020 |
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Dec 10, 2020 18:56
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Puzzle 86: Statement 2-1 cannot be true, since 2-2 would also be true in that case. That means 2-1 is false and and 2-2 is true. Thus both 1 and 4 are innocent. That means 1-2 is true and Farmhand 1 is in love with Farmhand 2. 4-1 is true, meaning 4-2 is false and Farmhand 3 is guilty. Checking Farmhand 3's statements, 3-1 is false and 3-2 is true.
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Dec 12, 2020 14:57
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Puzzle 88 : If A is guilty, A-1 is false, B-2 is false, and C-2 is true. That means C-3 is false => B-3 is false => A-3 is true. But B-1 is also false, since C-1 is false, and B has no true statements. If B is guilty, A-1 is true, B-2 is true => B-3 is false => A-3 is true, another contradiction. If C is guilty, A-1 is false, B-2 is false, and C-2 is false. If C-3 is true, B-3 is true => A-3 is false, and A-2 is true. B-1 and C-1 are false and A & C have met multiple times. If C-3 is false, B-3 is false => A-3 is true. A-2 is false, so A & C have met only once. Either way, C is guilty.
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Dec 15, 2020 14:19
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#91 Initially I had an overly complicated answer here that ignored the fact that all four were on duty during the thefts. Here's the simpler one. From the facts given (all were on duty during the thefts), we can get the first line below. Additionally, C1 or C2 cannot both be true, which means D2 is false. pre:A 1 2 3 B 1 2 3 C 1 2 3 D 1 2 3 ? F ? ? T ? ? ? ? T F ? F F F T T T T F T T F F Kangra fucked around with this message at 20:11 on Jan 3, 2021 |
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Jan 3, 2021 18:05
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There's something wrong with this problem. Even if you're extremely charitable and allow statements like "x is not as old as y' to mean 'x is younger than y', you still don't uniquely determine the jobs. All you can get is that the youngest, E, is the Synthetic Food Nutritionist and a few minor constraints such as only A or D can be the Lunar Energy Engineer e: My best guess for the intended solution: A: Comm Con B: Airfoil Tech C:Space Planner D: Lunar Energy Engr. E: Synth Food Nut. vvvvvvv The issue is that the negation in statement 1 is inherently ambiguous - but I suppose that if one assumes a solution does exist, then the interpretation that leads to a unique solution is the correct one (which is how I got my solution). Kangra fucked around with this message at 06:57 on Jan 12, 2021 |
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Jan 11, 2021 23:30
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#94 collage painting - D, 4th place holography - B, 2nd place laser etchings - C, 1st place reconstituted materials sculpture - A, 3rd place By #2 & #4, 4th-place must be the collage. By #3, only B or D entered the collage, so by #4, D must be 4th-place with the collage. By #5, A must be the third-place winner. By #6 and #4, A did not enter the laser etching, and neither did B, so C entered the laser etching. By #6, C cannot be second-place, so must be first-place, and then B is second-place. By #1, A did not enter the holograph, and thus must have entered the sculpture, leaving B to enter the holograph.
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Jan 12, 2021 20:26
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#95 A: 0 vowels, 6 syllables B: 1 vowel, 5 syllables C: 0 vowels, 7 syllables D: 2 vowels, 8 syllables #1,#2, and #6 tell us only D can have two vowel sounds. #5 & #6 let us know D cannot be the 5-,6-, or 7-syllable name. By #4, A must have zero vowel sounds, since if A had one, B & C would both be zero. By #6 and #2, only A can have 6 syllables, and that means only B can have 5 syllables. That means only C's name can have 7 syllables, and by #3 (and #5), it thus has zero vowel sounds, and B has one vowel sound. While D's name could have 8, or anywhere from 1 to 4 syllables, it's implied by the problem that it has 8.
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Jan 14, 2021 19:56
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#98 : A - Music, 0.5 SP B - Gulf, 0.6 SP C - Reading, 0.9 SP D - Boating, 0.75 By 1 & 4, B plays gulf. By 6 & 7, B must spend either .75 or 0.6 SP. By 2, 3, & 5, the boater spends 0.75, meaning B spends 0.6. By this & 3, the reader spends 0.9 SP, leaving the musician with 0.5. By 1, D must be the boater. By 1 & 4, A must be the musician, and C is the reader.
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Jan 21, 2021 13:55
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I know it didn't ask, but how does the astronaut in #99 know the Martian's not lying?
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Jan 24, 2021 05:09
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#103 Doman is an Uti. Aken's statement of not being an Uti would be truthful from a Yomi, and false from an Uti; Aken is necessarily a Grundi. (Also their second statement is false, since there must be one Uti in the group, and thus only one truthful statement about Doman.) Since Cwos cannot be a Grundi, their first statement would be true from a Yomi, an impossiblity. That means Cwos must be of the Uti, and so is Doman.
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Jan 25, 2021 08:45
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#105: I think the astronaut would simply give up. Who would farfels bear, to groan and sweat onto a weary spaceship, but that the dread of garbles eating them, puzzles the will, and makes us rather bear those plants we have, than fly with those we know not of? Thus logic does make cowards of us all.
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Feb 2, 2021 06:52
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#110 Jan Robinson By #1, Robinson is a Yomi. By #5, the Martian named after the biochemist is a Grundi, meaning Robinson is not the biochemist. By #6, Jan Robinson beat the engineer, so unless it was a game against themself, Robinson must be the pilot.
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Feb 8, 2021 19:29
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#112 Assumptions: The lettered statements each refer to unique individuals; that is, for each person, only one of the statements applies. Aken is not the person referred to as "Aken's friend". For seating descriptions, "opposite" is identical to "directly opposite". Likewise, "right", "left", or "between" imply a distance of fewer than four places. In Statement 4, the description of Mun indicates they are to be identified with statement c. To avoid having to say "the person identified in c", I'll just use those letters to refer to that individual. The question, then, is who is d? Aken - No (assumption that the two are not the same) Bal - No (By #5, e is opposite Bal, and by #1, g is opposite d) Mun - No (assumption of uniqueness, Mun is c by #4) Mark - No (By #1, also by uniqueness since Mark is g) Wora - No (By #2) Jones - No (By #6, Jones cannot be opposite Mark/g) Rider - Yes (especially bolstered if Jones is to the immediate right of Mark) Smith - No (By #4)
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Feb 15, 2021 20:45
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I rather liked the themed connections between the groupings, as it made them feel like part of a consistent system or story. Although it did always seem like there was one or two in the groups that didn't fit or made odd assumptions. The weighing and water puzzles were probably the best. The last set was possibly the weakest. It was often vague with wording, while also being mostly easy puzzles.
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Feb 16, 2021 23:12
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