A really cool combinatorics problem / puzzle Problem statement  6 scientists are working on a secret project. They wish to lock up the documents in a cabinet so that cabinet can be opened when and only when 3 or more scientists are present. What is smallest number of locks needed? What is smallest number of keys each scientist must carry??  Let $n$ be the number of locks. Let us name each scientist as $1,2,3,4,5,6$.Let $S_i$ be the set of keys that scientist $i$ is carrying and each scientist carries $k$ number of keys.we would get following results  $|S_i|=k$ $|S_p \cup S_q \cup S_r|=n$ $n=3k-(|S_p \cap S_q|+|S_p \cap S_r|+|S_r \cap S_q|)+|S_p \cap S_q \cap S_r|$  I tried manipulating the above results in hope of finding solutions but failed.Is there anything I am missing or is this approach entirely wrong?
Combinatorics riddle: keys and a safe There are 8 crew members, The leading member wants that only a crew of 5 people or more could open a safe he bought, To each member he gave equal amount of keys, and locked the safe with several locks.  What's the minimal number of locks he should put:   at least 5 crew members would be required to open the safe? Any team of 5 crew members could open the safe, but none of the 4 team crew members.   1,2 stands individually of course.
I am extremely poor in solving a problem quickly, which leaves me doing badly in olympiads and other contests. But, if I just don’t give up on a problem, no matter how much time it takes, I usually end up solving it. To have a future in mathematics or physics, is it necessary for me to excel at contests, because these days children excelling at contests have an advantage over those who don’t (not complete advantage, but advantage nonetheless)? Is  && statement() the same as if() statement()? How can I iterate over each file in a directory using a for loop?  And how could I tell if a certain entry is a directory or if it's just a file? I want to play a game in an older update, but can't see a way to choose a specific version. Is this possible? Comparing two java.util.Dates to see if they are in the same day Is there an idiom for people who boast too much? Is there any way to import a .json model into Blender? I only have access to the server via a terminal and I can't use graphical tools such as GParted! I want to create a new partition from a part of the root (about 768mb) for swap.  # df -h Filesystem      Size  Used Avail Use% Mounted on /dev/vda         20G  1.3G   18G   7% / udev             10M     0   10M   0% /dev tmpfs           199M  4.9M  194M   3% /run tmpfs           100M   12K  100M   1% /run/user tmpfs           5.0M  4.0K  5.0M   1% /run/lock Is there a way to replicate Nikon D3100 live view on my laptop screen? I own a Nikon D3100 and I need a way to see the live view of my camera on my laptop screen.  There must definitely be a way (I am not just aware of it). I have searched all around the web sites... I could not get a solution.  The Nikon Camera Control Pro 2 does not support D3100, and "Camera Control 5.1" connects and shoots remotely from a laptop but does not show live-view nor shoot in RAW.  There is a way of connecting my camera to a TV or monitor directly via an HDMI cable and getting the live view, but I want it displayed to my laptop screen. Prove that there exist two integers such that i - j is divisible by n. Here's the full question:  Prove that, for any $n + 1$ integers, $\{x_0, x_1, x_2, . . . , x_n\}$, there exist two integers $x_i$ and $x_j$ with $i \neq j$ such that $x_i − x_j$ is divisible by $n$.  Now, the integers aren't necessarily consecutive, positive, or without repeats. I've tried breaking this into cases, such as "$x_i = x_j, \dfrac{0}{n} = 0$" etc. But I don't think it's possible to exhaust the cases.  I feel like there should be an easy way to contradict this one, by saying that $x_i - x_j$ isn't divisible by $n$. But I just wasn't getting anywhere with that one.  I thought about using induction, but I would have just as hard a time proving this for $n$ integers before $n+1$ integers, so I don't think it would be possible for me to do it that way.  If anyone could help out, that would be wonderful. I'm pretty stuck here. How to unlock gungeoneers I've seen on a wiki that there is a ghost gungeoneer and a robot gungeoneer but it didn't say how to unlock them, so how do I unlock more gungeoneers ? Smooth option sometimes produces incorrect arrow tips in PGFplots It seems that the smooth option applied to lines sometimes effects the arrow tips and produces incorrect results.  No problem with vertical lines, but any other seems to have a problem. Is this a known bug, or is there a reason why the smooth should not be applied to straight lines?  Updated: to show problem does NOT occur with TikZ, and problem also occurs with curves if smooth option is applied.  \documentclass{article} \usepackage{pgfplots} \begin{document}  \tikzstyle{MyStyle}      =[-&gt;, black, ultra thick,       ] \tikzstyle{MyStyleSmooth}=[-&gt;, black, ultra thick, smooth]  \begin{tikzpicture} \begin{axis}     % 'smooth' option seems to have no effect on lines '1' and '2'     \addplot [MyStyle      , green ] coordinates{(0,0) (0,1)} node [above] {1};     \addplot [MyStyleSmooth, blue  ] coordinates{(1,0) (1,1)} node [below] {2};     %     % But, can't seem to have the 'smooth' on lines, '4', '5' and '6'     \addplot [MyStyle      , orange] coordinates{(2,0)   (3,0)   } node [right] {3};     \addplot [MyStyleSmooth, red   ] coordinates{(4,0)   (2,0.5) } node [below] {4};     \addplot [MyStyleSmooth, red   ] coordinates{(2,1)   (3,1)   } node [left ] {5};     \addplot [MyStyleSmooth, red   ] coordinates{(3,0.8) (2,0.8) } node [right] {6};      % Similar problem with curves: "non-smooth" is ok, "smooth" is not     \addplot[MyStyle,       mark=none, domain=0:1,samples=50, blue] (x,x*x+1);     \addplot[MyStyleSmooth, mark=none, domain=0:1,samples=50, red ] (x,x*x+2);  \end{axis} \end{tikzpicture}  \begin{tikzpicture} % These are all fine     \draw [MyStyle      , green ] (0,0)   -- (0,1)    node [above] {1};     \draw [MyStyleSmooth, blue  ] (1,0)   -- (1,1)    node [above] {2};     %     \draw [MyStyle      , orange] (2,0)   -- (3,0)    node [right] {3};     \draw [MyStyleSmooth, red   ] (4,0)   -- (2,0.5)  node [below] {4};     \draw [MyStyleSmooth, red   ] (2,1)   -- (3,1)    node [right] {5};     \draw [MyStyleSmooth, red   ] (3,0.8) -- (2,0.8)  node [left ] {6}; \end{tikzpicture} \end{document} Continued fraction of $e^{-2\pi n}$ I found this problem on a well-known problem solving website. It is apparently from Ramanujan.     With   $$\LARGE{a = \frac{1}{1 + \frac{e^{-2\pi}}{1 + \frac{e^{-4\pi}}{1 + \ddots}}}},$$   what is $a$?   The basic methods I know for questions of this type were not helpful to me.  I do not require a full solution, but feel free to post one if you would like. Hints or lecture notes on how to solve problems of this nature would be appreciated.   Keep in mind that I am at an undergraduate level in mathematics. There's a fairly standard two or three-semester curriculum for introductory quantum field theory, which covers topics such as:   classical field theory background canonical quantization, path integrals the Dirac field quantum electrodynamics computing $S$-matrix elements in perturbation theory, decay rates, cross sections renormalization at one loop Yang-Mills theory spontaneous symmetry breaking the Standard Model   What is a good, complete and comprehensive book that covers topics such as these? Every word is marked as misspelled in TeXstudio How can I tell if an AC adapter is compatible with my laptop? Possible Duplicate:        How can I tell if an AC adapter is compatible with my laptop?  For example, does the voltage output have to bee the same, or just close?  What power measurements matter? I have this question :  Proof if $f$ continuous in $x_0$ then there is a neighbourhood of $x_0$ so $f$ bounded.  I want to know if my proof is valid :  If continuous in $x_0$ then :  $$\lim_{x \to x_0} f(x)= f(x_0)$$  Therefore :  All $\epsilon&gt;0$ exists $\delta&gt;0$ so any $x$ that implies $|x-x_0|&lt;\delta$ implies $|f(x)-f(x_0)|&lt;\epsilon$.  Therefore for $\epsilon=1$ there is $\delta&gt;0$ so any $x$ that implies $|x-x_0|&lt;\delta$ implies $|f(x)-f(x_0)|&lt;1$.  $|x-x_0|&lt;\delta \rightarrow x_0-\delta&lt;x&lt;x_0+\delta$.  Lets choose $x_1,x_2$ that implies $x_0-\delta&lt;x_1&lt;x_2&lt;x_0+\delta$  Therefore in $[x_1,x_2]$ from Weierstrass from theorem bounded there. Bluetooth earphones work fine until sleep. After resuming from sleep however, they appear to connect for a brief moment before disconnecting. On blueman, the error given is Resource temporarily unavailable. This issue arose only after updating to 18.04 LTS.   Here's the terminal output for lsusb:  Bus 001 Device 002: ID 8087:8001 Intel Corp.  Bus 001 Device 001: ID 1d6b:0002 Linux Foundation 2.0 root hub Bus 003 Device 001: ID 1d6b:0003 Linux Foundation 3.0 root hub Bus 002 Device 004: ID 1bcf:0002 Sunplus Innovation Technology Inc.  Bus 002 Device 003: ID 04f2:b477 Chicony Electronics Co., Ltd  Bus 002 Device 002: ID 0a5c:21f1 Broadcom Corp. HP Portable Bumble Bee Bus 002 Device 001: ID 1d6b:0002 Linux Foundation 2.0 root hub What's the most efficient way of determine individual card values for my collection? I have several input and option elements on my page, each (well almost) have an event attached to update some text on the page once they change. I use jQuery which is really really cool :)   I also use Microsofts  framework, utilizing the UpdatePanel. The reason why I do that is that certain elements are created on the page based on some server-side logic. I don't really want to explain why I use the UpdatePanel - even if it could (it can with quite some effort) be rewritten to use only  I still want that UpdatePanel.  You probably guessed it - once I have a postback on the UpdatePanel the jQuery events stops working. I actually was expecting this, since the "postback" is not really a new postback so my code in document.ready that binds the events won't be fired again. I also confirmed my suspicion by reading up on it in the jQuery help libraries.  Anyway I'm left with the problem of rebinding my controls after the UpdatePanel is done updating the . I preferably need a solution that does not require adding more .js files (jQuery plug-ins) to the page but something as simple as being able to catch the UpdatePanel's 'afterupdating' where I can just call my method to rebind all the form elements.