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Everything posted by EventHorizon
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Happy Pythagorean Triple Day!! Also, Mother's Day.
EventHorizon posted a question in New Logic/Math Puzzles
Yes, I know Mother's Day is the much bigger deal (Happy Mother's Day to all mothers out there! I hope you feel loved and appreciated every day of the year.). However, today's date is a Pythagorean triple (5/13/12, 52+122=132), so I thought I'd post some math questions regarding Pythagorean triples. But first, here are some definitions to those who don't know what Pythagorean triples are: The Pythagorean theorem states that for a right triangle whose sides are of length a, b, and c where c is the longest length (the one opposite the right angle), a2+b2=c2. If all three of a,b, and c are positive integers, then {a,b,c} is a Pythagorean triple. Now, in honor of Mother's Day I'll call a Pythagorean triple a mother triple if there is no d greater than 1 such that {a/d, b/d, c/d} is also a Pythagorean triple (equivalently, at least 2 of {a,b,c} are relatively prime). Otherwise, it is a child triple. Here are the questions: 1. Let the date be of the form xx/yy/zz (ie, two digit year, any can be c (the longest)). What are the next 3 days that are pythagorean triples? 2. How many dates are mother triples given the two digit year? How many total dates are Pythagorean triples? 3. How many total dates are triples given a four digit year? 4. What was the last date that was a pythagorean triple given a four digit year? -
Amiab is right. On a side note, that is close to question 2 from my modifications I linked to earlier (though I used an analog clock as my example... so 12 holes).
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(by me ) (fyi, the answer to 4 ended up pretty interesting)
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I'll venture an initial guess for a good number of rounds to be twice the number of players, and I base this guess on absolutely nothing. Though players+1 seems like a good guess too. Just curious, when would you call the decks randomized? It seems you could get to the state where every situation could occur, but they wouldn't all have the same probability of occurring (and it may be impossible to reach uniformly random regardless of the number of rounds). That said, I'm thinking something like either 1/e or 1-(1/e) will be the optimal proportion of cards to pass (or maybe something like 1-(1/(e^p)) where p is the number of players). Could the optimal proportion of cards change for each round? Like it is best to pass 1/2 the first time, then 1/3 etc. Actually, it's hard to see a better proportion than 1/2 for 2 players as it maximizes the entropy. I'll think about this further... Great Post
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Well, if any two of {A,B,C} have the same prime factor, then the other must (trivial to prove). So we are looking for a case where all 3 are relatively prime. I'm thinking my approach will involve lots of modular arithmetic, but nothing of note to report right now. And given there's a prize on this one, I'm thinking it's been tried before by people more familiar with the necessary maths to solve it (I know next to nothing about rings/fields/etc). Still interesting though, so I'll play with it for a while.
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You misunderstood the method of choosing the elements of the column marked '0' It is not simply the fibonacci numbers plus one. For instance, the row starting with 4 would require you to find the 4 in the columns with positive integers. It occurs in the column marked '1' and in the row beginning with 1. To the right of that is a 7, so adding 1 to that makes it an 8, not 9 (the fibonacci+1). The next number on that row happens to be the 12 that wouldn't occur anywhere if choosing fib+1.
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Yup, that's the easy way I hinted at.
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Nana7 and rasar619 have the question right (a trillion is an insignificant fraction compared to the potential gold from the box), but both are low on the maximum payout. And as for the time limit, 24 hours was much more than is needed...even to max out the box. Here are two new questions... 1. What is the maximum amount of gold you could get out of the box? (in the form "5*(2^X)" dollars worth) 2. What is the shortest amount of time allowed to repay the debt that is still manageable? (Assume 1 second per button press, 3 seconds per button hold, and ignore any travel time) (FYI, I know 1 but haven't thought about 2)
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You are approached by a man holding a strange looking box. The box has 6 chambers numbered 1 through 6. Each chamber contains 1 token. Each chamber also has two buttons on it with labels A and B. Button A will remove a token from the chamber and add two tokens to the next higher numbered chamber (if it exists). Button B will remove a token from the chamber and swap the contents of the next two higher numbered chambers (if they exist). Chamber 6 has an extra button, C, which will remove a token from chamber 6 and cause the box to magically produce a dollar's worth of gold. If held down for 3 seconds, a button will perform its function repeatedly and instantaneously until there are no more tokens in the chamber (or up to an amount you are thinking of at that moment). Tokens will resize themselves so you don't need to worry about overcrowding a chamber. Any tampering with the machine beyond pressing the buttons will cause it to shrink out of existence. The man wants to sell you the box for one trillion dollars. You happen to be able to take out the loan to buy it, but only though some shady people who'll kill you if you don't pay back the loan within 24 hours (they will leave you alone if you pay it back and they'll take gold). Do you buy the box?
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It's not ebay, simply because eBay exists and the person who made this contest wouldn't name his new company eBay. From the contest page, someone has won the $10,000. Once they release more information, I'll update the thread. I'd like a copy of the key so I can try and solve the puzzle myself. Also, where the key was found would be interesting.
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Source page with contest info Solve this system of equations and find the "key" to discover the new company name. Be the first to find it and you win $10,000. Also, to get the $10,000 you need to sign a nondisclosure agreement. So if you do find the answer... I guess you shouldn't post it.
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Hey Braindenizens. The posts on seem to have stopped, so I thought I would simplify it a little to try and find an eventual solution. 1. Given a well shuffled deck of 8 cards (numbered 1 through 8), what is the probability that when the cards are dealt out that no card's number will reflect the order it was dealt? (i.e., card labelled 1 is not dealt first, card labelled 2 is not dealt second, etc) 2. What is the probability in terms of N, where N is now the number of cards in the deck? 3. What is the probability if we use an infinite deck of cards? (yes, take the limit as N approaches infinity)
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Agreed, it's all downhill after you get a theme song.
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EventHorizon, started well, but perhaps swayed somewhere in between. See the Spoiler.