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Tax Refund posted: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? This is the kind of puzzle I can get behind. By which I mean I believe I have the solution. Vend one bar from the first machine, two bars from the second, and so on up to all hundred bars from the hundredth machine. Put all of this chocolate on the scale. If all of the bars weighed 100 grams, I believe the total weight should be 505,000 grams (100 * 101 / 2 * 100), but it will weigh more than that. Call the total weight W, and calculate n = W - 505000. This is the number of grams of excess weight, and thus the number of bars that each weigh one gram too much. Since we took n bars from machine number n and from no other machine, the nth machine is the one that has the overweight bars. It's not hard to prove that the problem is impossible to solve in fewer weighings than one, but I'll leave that proof as an exercise for the reader.
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Jan 24, 2015 14:58
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| # ¿ May 22, 2024 18:54 |
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That's a point... I keep forgetting that it's possible to do puzzles without math sometimes.
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Jan 26, 2015 02:20
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That's perfect. Why haven't you posted in my Folklore thread so I can put your name in the credits? You're the first boss of the game! I should probably post a puzzle here as well, while I'm at it. This is a topic that I personally find fascinating but doubt anybody else cares about. Puzzle TR02: Suppose that you arrange the numbers from One to One Billion in alphabetical order. The words "a" and "and" do not appear anywhere in this list. What is the number immediately preceding Five? Solution: Fifty-two thousand, two hundred two. Plus some extra words to disguise the actual length of the answer. You know how it goes. Nidoking fucked around with this message at 00:29 on Jan 28, 2015 |
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Jan 26, 2015 03:07
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Roonerspism posted:I think it's actually I think everyone's had enough time to think of numbers, so I'm going to say this is the answer I came up with. I can't absolutely guarantee its correctness because there are a billion numbers and I've never actually sat down and alphabetized them, but after years of pondering this problem, that's the number I came up with. There ARE numbers between that and Five, but they're well over one billion. (The first such number I can think of is Fifty-two trillion, rather obviously.) There's definitely a temptation to go for longer numbers, but length is only relevant to alphabetical order when the shorter word matches the beginning of the longer word.
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Jan 28, 2015 00:28
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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.
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Feb 1, 2015 15:41
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| # ¿ May 22, 2024 18:54 |
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Air is lava! posted:Technically, you could argue that thisv question is just wrong. The two different x come from different contexts, so they can attain different values. That's just abuse of notation. Sometimes you actually see terrible stuff like that if an author of a paper is a little sloppy. This reminds me of a story from my algebra class. One of the students, for some reason, always wrote his 'x's so that there was a tail on the upper right, like he was slow to lift his pencil from the paper once he was finished writing, except the tails were far too consistent for it to have been accidental. The teacher once called him out for it, in the fashion that teacher was famous for. He was discussing a problem where there were both M and m as variables, and many people were confused because they didn't understand that those were distinct. So he wrote them both on the board. "This is a capital M, this is a lowercase m. Do they look anything alike? No, those are different variables. Here's a capital A, and here's a lowercase a. Completely different. Here's a capital X, here's a lowercase x, and here's a Tariq x~. Not the same thing at all." He was one of my favorite teachers, and I'm sad that I was sitting in the back of the room doing independent study instead of taking that class.
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Mar 7, 2015 12:45
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