| Is CP a chess genius? =O And is the riddle officially solved now? |
| I'm gonna re-use a riddle from when I hosted The Decision Game; Pantry You and your teammates wake up inside a small storage room full of boxes stocked with potatoes, canned food, crackers and other non-perishables. The door is locked and you need a two digit pincode to open it. Right next to the door is a small note that says "faces times edges". You search the room for more clues and find 2 boxes that are not square like the rest, but instead have very odd shapes. One of them is a tetrahedron and one is a pyramid, both with equal side lengths of 1 meter. The two have been tapped together on a triangular face with the vertices aligned, but how many faces and edges does this shape have? Enter the correct pincode to unlock the door and escape! |
| Enters 45 as pin code |
| @grave_robber Correct :) ... But how did you know to subtract 2 extra faces and 2 extra edges? ;) |
Zymf said: @grave_robber Correct :) ... But how did you know to subtract 2 extra faces and 2 extra edges? ;) I wasn't sure what the shape would look like so I googled "tetrahedron and pyramid connected by triangle face" and I got this image result: http://tardus.net/pyramidPuzzle/2tetrasJoined.jpg So I drew it and counted manually: |
| More shape puzzles please, this was fun (almost made some real tetrahedron and pyramid cutouts to do it XD) and if anyone feels like solving this, you can: Puzzle said: You have ten identical bags of identical coins – except that one bag contains counterfeit coins that weigh one gram less than real coins. The difference isn't enough for you to determine which bag has the bad coins by feel; you need to weigh them on your highly accurate scale. You know the weight of a good coin (20 grams), and your scale can easily differentiate weighs down to a fraction of a gram. If you are only allowed one weighing, how do you determine which bag has the counterfeits? |
grave_robber said: Ooh I think I got it. You take 1 coin from one bag, 2 from the next bag, and so on, until you take 10 from the tenth bag, then weigh all the coins together. The then subtract the total weight from 55 (which would be the total weight with no counterfeights), and that will tell you which # bag is the fishy one.More shape puzzles please, this was fun (almost made some real tetrahedron and pyramid cutouts to do it XD) and if anyone feels like solving this, you can: Puzzle said: You have ten identical bags of identical coins – except that one bag contains counterfeit coins that weigh one gram less than real coins. The difference isn't enough for you to determine which bag has the bad coins by feel; you need to weigh them on your highly accurate scale. You know the weight of a good coin (20 grams), and your scale can easily differentiate weighs down to a fraction of a gram. If you are only allowed one weighing, how do you determine which bag has the counterfeits? |
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grave_robber said: @Togs Perfect =3 @Zymf, I thought it only worked because the lengths are the same (1) and thus the angles would be the same and it would be aligned. Btw, your drawing is way better than mine... I hope mine still made sense ^^' I wouldn't call it better : D. the resulted figure should be triangle oblique prism with faces 3 and 5 being parallel and they look nothing like parallel in that picture. it should look like this:  which is basically what you draw grave but without the cut to separate the two peaces. In other words, Grave: 1, Google: 0. |
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| Ran into an interesting problem at work today and had some fun solving it. A boy take a whole bunch of identical little cubes and stacks them all together to create one larger cube. He then paints some of the sides of the large cube to make it look nice, but his sister comes in and accidentally knocks the whole thing down. After being knocked down, 45 of the little cubes did not have any paint on them. How many faces of the large cube did the boy paint originally ? |
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| Hmm... interesting :D Let's see - The number of cubes could be any cube number (n^3): 1^3 = 1 2^3 = 8 3^3 = 27 4^3 = 64 5^3 = 125 6^3 = 216 Everything below n=3 result in less than 45 small cubes. So the large cube is at least 4x4x4 How many cubes can we subtract from each of these numbers, if the boy painted all 6 sides of the large cube? From a large cube with length n, the maximum number of small cubes with paint would be: n*n*2 + n*(n-2)*2 + (n-2)*(n-2)*2 (only works with n>2 though) 3x3x3 => 3*3*2 + 3*1*2 + 1*1*2 = 26 4x4x4 => 4*4*2 + 4*2*2 + 2*2*2 = 56 5x5x5 => 5*5*2 + 5*3*2 + 3*3*2 = 98 6x6x6 => 6*6*2 + 6*4*2 + 4*4*2 = 152 If we subtract ALL the outer cubes from the 6x6x6 we are left with 216-152=64, which surpasses 45 So the large cube is at most a 5x5x5 Now with a 5x5x5 or a 4x4x4 cube, how many sides did the boy paint. Hmmm... 4x4x4 cube has 64 small cubes. Subtract one side (4*4=16) and you get 48. You can't paint any more sides without surpassing 45. It can't be the 4x4x4 5x5x5 cube has 125 small cubes. Subtract one side (5*5=25) and you get 100. Even if you paint the opposite side, you would only get down to 75, so at least one of the adjacent sides MUST be painted as well. Subtract one adjacent side (4*5=20) and you get 80.... Waiiiiit a second - It's just 3*3*5=45, which means that 3 of the sides on the 5x5x5 must have been painted. @Togs |
ZymfMar 20, 2018 6:04 AM
| Hmm, this seems near impossible with how my mind works. Considering the boy cannot paint the inside of the large cube, and any one of those 45 little cubes could have paint on one of its side and be facing down. Though, I look forward to seeing how it's solved. |
AstrosMar 20, 2018 10:13 AM
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Zymf said: Good thought but this isn’t quite the answer ! 3x3x5 is indeed the right amount of cubes but that doesn’t imply 3 faces..5x5x5 cube has 125 small cubes. Subtract one side (5*5=25) and you get 100. Even if you paint the opposite side, you would only get down to 75, so at least one of the adjacent sides MUST be painted as well. Subtract one adjacent side (4*5=20) and you get 80.... Waiiiiit a second - It's just 3*3*5=45, which means that 3 of the sides on the 5x5x5 must have been painted. |
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Zymf said: You got it ! The way I approached this problem was using the fact that every cube contains a smaller cube with a side length of 2 less inside, which gets you the answer quite quickly if you build off of that. It's not? *Tries harder to mentally visualize* ... Oh yea, you're right x.x The boy painted 4 faces all the way around leaving two opposite sides un-painted. Brainfart.. @Togs |
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| Cunning Chris decided to go to his local pub one night in the rain as he was bored at home. As he entered the pub he greeted some old acquaintances, like Jazz Jake and Swift-Hand Steve. He purchased a drink and sat with his old friends and had an idea how to lighten up the evening. "I have a challenge for you Swift-Hand Steve" said Cunning Chris. "Well bring it on then Cunning Chris" goaded Steve, forgetting how cunning Cunning Chris was. Chris then grabbed a handful of poker chips from behind the bar, scattered them across the table and placed two paper cups by opposite edges. "Right, here we have fifty white chips and fifty black chips. My challenge is that I can place all the black chips in my cup before you can place all the white chips in your cup. The rules are, you must have one hand on the cup at all times, you can only touch one chip at a time, you can only use your hands and you can't use any object to help you gather the chips." "Easy" replied Swift-Hand Steve. A smile had spread across his face, he knew he had the quickest hands in town. "Ok then, Go" Swift-Hand Steve was true to his name and he quickly had over ten chips in his cup before Cunning Chris even had two. But moments later, it was Cunning Chris who wore the giant grin. What had he done to secure his victory? |
Togs said: Exactly :D@Zymf Chris never specified that you were only allowed to pick up chips of your own color, so maybe he grabbed a black chip first ? So when Steve "finished" he only had 49 chips without realizing it. He hid one of the black chips underneath his cup. After they had both collected all the other chips, he liftet up his cup to reveal his victory ;) |
| Jimmy needs to calculate (a+b)/c for a homework assignment where a, b, and c are positive integers. He first types in a+b÷c directly into his calculator and gets 11 as a result. He then tries b+a÷c and gets 14. Jimmy then realizes he forgot his parentheses, so he enters (a+b)÷c next and ends up with the correct answer. What was the correct answer ? |
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| I have no idea :) |
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