Can't install rmagic due to the dependencies errors
How do I resolve unmet dependencies after adding a PPA?
How to get memory usage under Windows in C++ I asked  yesterday, approximately 18 hours ago ("2011-12-13 20:24:03Z" according to the hover text), and it's still being displayed as "today".  I assume the "Z" in the timestamp refers to Zulu time, so I , and the date (2011-12-14) differs from the timestamp on the question.  I don't pretend to understand all of the complexities when converting time zones, but I thought it was strange enough to report. If it matters at all, I'm in US-Eastern. Get latest Flash player on Ubuntu Please tell me which choice below (or rather a better choice that I need to know) works better in my sentences in informal English:     I got myself out on a limb. I need to clear my check by tomorrow. No matter how, I must ............ 2 million bucks by tomorrow.   a) line up [to prepare, organize, or arrange something]  b) hook up [hook somebody up with something to help someone get something that they need or want]  c) fix up [to arrange for someone to have something that they need]  For me, all three options work in this sense with the same meaning and informality degree. How to convert string to byte array in Python Here's an example question from the :     On the last day of a one-week sale, customers numbered 149 through 201 were waited on. How many customers were waited on that day?   Possible answers: 51, 52, 53, 152, 153.   The correct answer here is 53, which is the result of (201-149) + 1 = 53.  What's the reasoning in adding the +1? How does one add boarding passes to Apple Wallet? We have $q, p$ prime and $tq = (p-1)(p+1)$ for some positive integer $t$. The book I'm reading says that because $q$ is prime, it divides $p-1$ or $p+1$.   Question: Why exactly does it matter that $q$ is prime here? Alternate Row Coloring in Django Template with More Than One Set of Rows Django templates offer the builtin tag cycle for alternating between several values at different points in a template (or for loop in a template) but this tag does not reset when it is accessed in a scope outside of the cycles definition.  I.e., if you have two or more lists in your template, the rows of all of which you'd like to use some css definitions odd and even, the first row of a list will pick up where the last left off, not with a fresh iteration from the choices (odd and even)  E.g., in the following code, if the first blog has an odd number of entries, then the first entry in a second blog will start as even, when I want it to start at odd.  {% for blog in blogs %}   {% for entry in blog.entries %}     &lt;div class="{% cycle 'odd' 'even' %}" id="{{entry.id}}"&gt;       {{entry.text}}     &lt;/div&gt;   {% endfor %} {% endfor %}   I've tried obviating this by patching with the resetcycle tag offered here:    to no avail.  (The code didn't work for me.)  I've also tried moving my inner loop into a custom tag, but this also did not work, perhaps because the compile/render cycle moves the loop back into the outer loop?  (Regardless of why, it didn't work for me.)  How can I accomplish this simple task!?  I'd prefer not to create a data structure in my view with this information pre-compiled; that seems unnecessary.  Thanks in advance. Let $d=GCD(n^2+5,n^3-5n^2+6n)$ show that $d|630$ How to make a different font and color of text in labels? Is it possible to make a bold or italic, and change the color of individual words Label?  Example: the first line is bold in red text, font 12, second row 10 green italic font Why does Microsoft Word 2010 show an error for the following sentence?     1.&nbsp;Where should this car be parked?   Word 2010 also suggests changing the sentence to     2.&nbsp;Where this car should be parked?   Is the sentence (1) wrong? If yes, why? What about sentence (2)? My /boot partition hit 100% and now I can't upgrade. Can't remove old kernels to make room My first issue was when I tried to apt-get update or apt-get upgrade. On upgrade I get the following error:  You might want to run 'apt-get -f install' to correct these. The following packages have unmet dependencies: linux-image-server : Depends: linux-image-3.2.0-27-generic but it is not installed E: Unmet dependencies. Try using -f.   I tried running apt-get install -f and this was the output(after saying yes on prompt)  (Reading database ... 186183 files and directories currently installed.) Unpacking linux-image-3.2.0-27-generic (from .../linux-image-3.2.0-27-generic_3.2.0-27.43_amd64.deb) ... Done. dpkg: error processing /var/cache/apt/archives/linux-image-3.2.0-27-generic_3.2.0-27.43_amd64.deb (--unpack):  failed in write on buffer copy for backend dpkg-deb during `./boot/System.map-3.2.0-27-generic': No space left on device  No apport report written because the error message indicates a disk full error                                                                           dpkg-deb:    error: subprocess paste was killed by signal (Broken pipe) Examining /etc/kernel/postrm.d . run-parts: executing /etc/kernel/postrm.d/initramfs-tools 3.2.0-27-generic   /boot/vmlinuz-3.2.0-27-generic run-parts: executing /etc/kernel/postrm.d/zz-update-grub 3.2.0-27-generic /boot/vmlinuz-3.2.0-27-generic Errors were encountered while processing: /var/cache/apt/archives/linux-image-3.2.0-27-generic_3.2.0-27.43_amd64.deb E: Sub-process /usr/bin/dpkg returned an error code (1)   I tried running apt-get autoremove and it gives me the same error as apt-get upgrade.  When I run df, I get this for /boot:  /dev/sda1                    233191     230297         0 100% /boot   So, I read elsewhere that I should try to purge old kernels. I checked for what kernels I had with:  $ dpkg -l linux-image-\* | grep ^ii ii  linux-image-2.6.38-13-server  2.6.38-13.52  Linux kernel image for version 2.6.38 on x86_64 ii  linux-image-3.0.0-13-server   3.0.0-13.22   Linux kernel image for version 3.0.0  on x86_64 ii  linux-image-3.0.0-14-server   3.0.0-14.23   Linux kernel image for version 3.0.0  on x86_64 ii  linux-image-3.0.0-15-server   3.0.0-15.26   Linux kernel image for version 3.0.0  on x86_64 ii  linux-image-3.0.0-16-server   3.0.0-16.29   Linux kernel image for version 3.0.0  on x86_64 ii  linux-image-3.0.0-17-server   3.0.0-17.30   Linux kernel image for version 3.0.0  on x86_64 ii  linux-image-3.2.0-24-generic  3.2.0-24.39   Linux kernel image for version 3.2.0  on 64 bit x86 SMP ii  linux-image-3.2.0-25-generic  3.2.0-25.40   Linux kernel image for version 3.2.0  on 64 bit x86 SMP ii  linux-image-3.2.0-26-generic  3.2.0-26.41   Linux kernel image for version 3.2.0  on 64 bit x86 SMP   When I try to remove the oldest with this:  $ sudo apt-get purge linux-image-2.6.38-13-server Reading package lists... Done Building dependency tree Reading state information... Done You might want to run 'apt-get -f install' to correct these: The following packages have unmet dependencies: linux-image-server : Depends: linux-image-3.2.0-27-generic but it is not going to be     installed E: Unmet dependencies. Try 'apt-get -f install' with no packages (or specify a solution).   How can I free up or extend boot without messing up my installation? I am looking for a YA book about a boy who learns to use powers from a mind-adept, in a seemingly abandoned house. The adept helps him open his mind by using a cube. One of his powers, I believe, is to go into dreams of people that  he knows. He helps his young female neighbor friend by doing this, opening a door in her mind. He tries to open the mind-door of a local older bully, but finds out the young man can't be helped because he has been abused mentally and physically. I don't believe the book I read is past 1995. It might have the word "adept" in it. date difference in days, in Android Running programs in the background from terminal I am reading the Dummit and Foote's proof of Cauchy's theorem for abelian groups. On the highlighted line, how do you conclude from $y\notin N$ and $y^p\in N$ that $\langle y^p \rangle \neq \langle y \rangle$ and that $|y^p|&lt;|y|$?     Proposition 21.   If $G$ is a finite abelian group and $p$ is a prime dividing $|G|$, then $G$ contains an element of order $p$.      Proof:   The proof proceeds by induction on $|G|$, namely, we assume the result is valid for every group whose order is strictly smaller than the order of $G$ and then prove the result valid for $G$ (this is sometimes referred to as complete induction).   Since $|G| &gt; 1$, there is an element $x \in G$ with $x \neq 1$.   If $|G| = p$ then $x$ has order $p$ by Lagrange’s Theorem and we are done.   We may therefore assume $|G| &gt; p$.      Suppose $p$ divides $|x|$ and write $|x| = pn$.   By Proposition 2.5(3), $|x^n| = p$, and again we have an element of order $p$.   We may therefore assume $p$ does not divides $|x|$.      Let $N = \langle x \rangle$.   Since $G$ is abelian, $N \trianglelefteq G$.   By Lagrange’s Theorem, $|G/N| = |G|/|N|$ and since $N \neq 1$, $|G/N| &lt; |G|$.   Since $p$ does not divide $|N|$, we must have $p \mid |G/N|$.   We can now apply the induction assumption to the smaller group $G/N$ to conclude it contains an element $\bar{y} = yN$, of order $p$.   Since $y \notin N$ ($\bar{y} \neq \bar{1}$) but $y^p \in N$ ($\bar{y}^p = \bar{1}$) we must have $\langle y^p \rangle \neq \langle y \rangle$, that is, $|y^p| &lt; |y|$.   Proposition 2.5(2) implies $p \mid |y|$.   We are now in the situation described in the preceding paragraph, so that argument again produces an element of order $p$.   The induction is complete.      (Original image .) Is the Ohm's law violated here? And also why is electricity generated at 11,000 V in India? I need to find limit points of the set $\{  \frac{1}{n} +\frac{1}{m} \mid n, m = 1,2,3,\dots \}$.  My try :   If both $m$ and $n$ tend to very large values say $\infty$ then the value of $\{ \frac{1}{n} +\frac{1}{m} \}$ tends to $0$, and if only one of $m$ or $n$ tends to very large values, then the set $\{ \frac{1}{n} \mid n=1,2,3,\dots\} $ acts as limit points.   So is it true that a set of limit points is $\{0\}$ $\cup$  $\{\frac{1}{k} \mid k=1,2,3,\dots \}$?  How should I write this proof rigorously? What happened to Saruman after he was knifed by Grima? Saruman was one of Maiar, like Gandalf. Meaning that he had a mortal body but presumably, when that body was killed, would simply return back to his ("spirit?") being.  Is there any confirmation from Tolkien that this is indeed what happened after he was knifed by Grima? Or was he somehow completely destroyed since he was de-powered by Gandalf?