I am trying to catch the legendary lake guardians, and have already caught Axelf and Mesprit.  I enter Nameless Cavern at exactly 20:00, but nothing happens. I look over the portal, but all it says is "A mysterious ring is floating in the air...".  How do I catch Uxie?
Every time since a few days ago after I caught Mespirit, when I go to the portal in the Nameless Cavern and I press A to interact. All that happens is a short dialog sequence and nothing else. I know the requirements are along the lines of having 3 max happiness Pokemon with you in your party which I do. I have also downloaded the recent update/patch from the eShop if that helps at all. How can I make Uxie and Azelf appear in the cavern?
In Windows, 'removable' drives cannot be partitioned, without the partitions not appearing and receiving drive letters, apart from the first partition.  Partitioned flash drives work fine in Linux.  What I want to do is partition a 8GB flash drive into two, with the first partition containing various boot disc images, which can be selected to boot from, and have various portable apps, documentation, utilities, scripts, etc, on the other partition.  I currently do this using two separate flash drives, but I want to be able to use a single flash drive for everything, which I keep on my keyring.  It is not possible to do this without having at least two partitions, as the ISO boot selector for the boot disc images will not work otherwise.  I know there is a method where you can flip a bit, using Lexar's USB Format Tool, which fools Windows into thinking that the partitions are on a fixed hard drive.  However, this apparently makes it difficult to eject the flash drive.  Flash drives with  from SanDisk had two partitions, presenting itself to the host system as a USB hub with a CD drive and standard USB mass storage device attached.  This is what I would like to emulate, if possible. If $a,b$ are positive integers such that $\gcd(a,b)=1$, then show that $\gcd(a+b, a-b)=1$ or $2$. and $\gcd(a^2+b^2, a^2-b^2)=1$ or $2 $ We all know you can't do the following because of ConcurrentModificationException:  for (Object i : l) {     if (condition(i)) {         l.remove(i);     } }   But this apparently works sometimes, but not always. Here's some specific code:  public static void main(String[] args) {     Collection&lt;Integer&gt; l = new ArrayList&lt;&gt;();      for (int i = 0; i &lt; 10; ++i) {         l.add(4);         l.add(5);         l.add(6);     }      for (int i : l) {         if (i == 5) {             l.remove(i);         }     }      System.out.println(l); }   This, of course, results in:  Exception in thread "main" java.util.ConcurrentModificationException   Even though multiple threads aren't doing it. Anyway.  What's the best solution to this problem? How can I remove an item from the collection in a loop without throwing this exception?  I'm also using an arbitrary Collection here, not necessarily an ArrayList, so you can't rely on get. Why don't applications update by calculating difference? How do I get the difference between two Dates in JavaScript? I'm creating an application which lets you define events with a time frame.  I want to automatically fill in the end date when the user selects or changes the start date.  I can't quite figure out, however, how to get the difference between the two times, and then how to create a new end Date using that difference. In the Harry Potter series is it clear that magically-inclined individuals (witches/wizards) will live to be considerably older than their Muggle counterparts? From the question , it is given that Dumbledore was roughly 115 years old when he died, and at that age he hardly seemed ready to kick the bucket naturally.  With that in mind, how old will a wizard typically live to be, barring accidents or murder? Let $\mathbf{Y}$ be a random vector. Are $k$th moments of $\mathbf{Y}$ considered? I am self-learning on linear model theory right now, and one thing I find surprising is that although $\mathbb{E}[\mathbf{Y}]$ is defined for a random vector $\mathbf{Y} = \begin{bmatrix} y_1 \\ y_2 \\ \vdots \\ y_n\end{bmatrix}$, there is no mention of further moments besides the covariance matrix.  Google searching hasn't turned up much. Are $k$th (raw) moments of $\mathbf{Y}$ considered, or is there a different idea I don't know about?  I am learning from the text  (the TOC starts in p. 17 of the linked file). By "considered," what I mean is is there such a thing as $\mathbb{E}\left[\mathbf{Y}^k\right]$, and if so, how would such a concept be defined? The book I have only covers the first raw moment, and I find it a bit strange that there is no mention of how to define $\mathbb{E}\left[\mathbf{Y}^k\right]$ given my experience in univariate probability, nor do I have the expertise to define it.  Furthermore, if $\mathbb{E}\left[\mathbf{Y}^k\right]$ isn't defined, is there perhaps a related concept that I don't know about that is used instead? WIFI connection not working I ran these commands:  sudo rmmod iwlwif sudo modprobe iwlwifi 11n_disable=1 sudo su echo "options iwlwifi 11n_disable=1" &gt;&gt; /etc/modprobe.d/iwlwifi.conf   Now, my WiFi connection is lost and there is no WiFi option in the network manager. How can I restore it? How to use wc and piping to find how many files and directories are in a certain directory? How can I use word counter (wc) and piping to count how many files or directories are in the /usr/bin directory? Order of products of elements in a finite Abelian group We want to show that if $a,b\in G$ where $G$ is a finite Abelian group, we have  $\operatorname{LCM}(|a|,|b|) = |ab|$ given that $ab \neq e$.   How I approached this question was by saying let $\operatorname{LCM}(|a|,|b|) = L$. Then if we show that $L$ divides $|ab|$ and $|ab|$ divides $L$ then $|ab| = L$. Showing that $|ab|$ divides $L$ was fine. But then I am having trouble with the second part that is showing $L$ divides $|ab|$. For this, so far I have:  Consider $(ab)^{|ab|} = e = a^{|ab|}b^{|ab|} $  since we know $a \neq b^{-1}$ by assumption then we can conclude that $|a|$ divides $|ab|$ and $|b|$ divides $|ab|$. We know that $L = \frac{|a||b|}{\operatorname{gcd}(|a|,|b|)}$ from here can we conclude that $L$ divides $|ab|$? Is it possible to play FIFA 2015, 4v4 on 2 Consoles? I've the following situation: My friends and I, we're 8 in total, want to play Fifa 15 4v4. I know that the Playstation 4 just supports 4 controllers and thats why we think about a 2 console + 2 TV's setup, so 4 players per console.   So my question is: Is it possible or is there a game mode available were we could play such a setup?  Update: After I got the first answer I've to add that I know that the only way to play on 2 consoles would be through the psn online features. So we did not plan to play on Lan. We just want to sit in the same room with 2 consoles and play against each other over psn. Should I compile release builds with debug info as "full" or "pdb-only"? Why does input capacity of an opamp increase noise gain? On the image below you can see an inverting amplifier with input capacity C1 and capacity C_F for general stability.     Now my book shows me following frequency plot:  A_n ... Noise-Gain  A_D ... open-loop gain  A_U ... V_out / V_in --> inverting amplifier gain     You can clearly see that C1 is increasing the noise gain at some point. And C_f is stabilizing A_n and A_U.   But why exactly does C1 increase the noise gain exactly? I changed my username and now I cannot play online I'm going through MSIL and noticing there are a lot of  instructions in the MSIL.  The MSDN article says they take no action and are used to fill space if the opcode is patched. They're used a lot more in debug builds than release builds.  I know that these kinds of statements are used in assembly languages to align later instructions, but why are MSIL nops needed in MSIL?  (Editor's note: the accepted answer is about machine-code NOPs, not MSIL/CIL NOPs which the question originally asked about.) In going through the review tool I often come across poor quality answers to poor quality questions. In particular, I will occasionally see questions that are simply asking for links to products, asking for tools, etc. Here are three examples that I've come across just this morning.        Clearly it's appropriate to go to the question and vote to close it (these are examples of not constructive) if they aren't already closed, but what should be done with the answers?    Should the answer be marked for deletion? Much of the normal reasoning for not providing link only answers is defeated if the question is simply asking for links. I'm tempted to mark them as 'looks good' for that reason, because it's the question that is flawed. Is that correct, or should I evaluate all answers as if they were responding to a quality question? Default Class Accessibility in C# by default is a class:   private ? internal ?  sealed ? Mobile site says "your answer couldn't.." when submitting a question edit that creates a new tag Noticed this on Bicycles, but I presume it affects any Stack Exchange site   Write question Edit question, write a new tag Submit edit   When performing the above steps, I properly get an error message, but it says "answer" instead of "question" or "edit". Speed of light in a gravitational field? How do I solve the speed of light in gravitational field?  Should I just add gravitational acceleration in speed of light?  $$c'=c_0+g(r)t~?$$ I would like to update VirtualBox on my 32 bit Ubuntu 15.04 machine from 4.3.26 to 4.3.28, using a Debian package I manually downloaded from . But simply double-clicking the package opens up the Software Centre which gives an error that installing it would break an existing package (the exact message was:   Breaks existing package 'virtualbox' conflict: virtualbox-4.3 ()   and it does not appear to have an option to ignore this and proceed with the installation anyway. Is there a command-line work-around this problem, perhaps?