Show that $24 \mid (n - 1)n(n + 1)(n + 2)$ for all positive integers $n$. Show that $24 \mid (n - 1)n(n + 1)(n + 2)$ for all positive integers $n$.  Since $24 = 2^3 \cdot 3$ it follows that $(n - 1)n(n + 1)(n + 2)$ is divisible by $3$ since it contains the product of three consecutive numbers. If $n$ is odd then $n-1$ and $n+1$ are even and other one of them is divisible by $2$ and the other by $4$, thus the product is divisible by $2^3$. If $n$ is even then $n$ and $n+2$ are both even and similar reasoning follows that other one is divisible by $2$ and the other by $4$. Is this correct? For the odd case if I have that $n=1$ then $n-1=0$ and $n+1=2$, but neither one is divisible by $4$ here? Or does $0$ count as being divisible by $4$?
Using the fact that $n^3 + 5n$ is divisible by $6$, infer that $n^4 - 2n^3 + 11n^2 + 14n$ is always divisible by $24$ The problem is as stated in the title. I've tried factorization, but it doesn't seem to yield anything. Any help is greatly appreciated. Also please don't use modular arithmetic, thank you.
What does it mean by 'bringing him gasping to all fours' in Harry potter? Immediately, his insides started writhing as though he'd just swallowed live snakes -- doubled up, he wondered whether he was going to be sick -- then a burning sensation spread rapidly form his stomach to the very ends of his fingers and toes -- next, bringing him gasping to all fours, came to a horrible melting feeling, as the skin all over his body bubbled like hot wax -- and before their eyes, his hands began to grow, the fingers thickened, the knuckles were bulging like bolts -- his shoulders stretched painfully and a prickling on his forehead told him that his hair was creeping down toward his eyebrows -- his robes ripped as his chest expanded like a barrel bursting its hoops -- his feet were agony in shoes four sizes too small      Harry potter and the chamber of secrets   It is the description of the feeling of drinking Polyjuice Potion, which makes someone into a different person, from the point of view of Harry Potter turning into Gregory Goyle.  And I don't understand what 'bring him gasping to all fours' means, especially 'four'. Does four here mean number 4? Conditionally Rendering Visualforce element based on user profile I have a controller extension and want to use the Tabbed view to get rid of the Chatter caused endless scrolling.      public class UserProfileGetter {     public UserProfileGetter(ApexPages.StandardController controller) {      String userProfileName = [select u.Profile.Name, id, UserType from User u where u.id = :Userinfo.getUserId()].Profile.Name;      system.debug('Profile===================='+userProfileName);      }    }   When I run the debug it shows my Profile as 'System Administrator' correctly.  My VF page code:  &lt;apex:page standardController="Account" extensions="UserProfileGetter" showHeader="true" tabStyle="account" &gt;   &lt;apex:tab label="Opportunities" name="Opportunities" id="tabOpp" rendered="{userProfileName!='Chatter Content'}"&gt;          &lt;apex:relatedList subject="{!account}" list="opportunities" /&gt;       &lt;/apex:tab&gt;        &lt;/apex:tabPanel&gt; &lt;/apex:page&gt;   So the logic for render I want is - if the user profile name isn't Chatter Content, render. But if I have this logic in the page it won't render for anyone, including me. Ideas? Thanks  Amending - to query other variable on the user, a checkbox called "Admin Team" to display the tab for Account Teams related list, I query for both like this:  public class UserProfileGetter {     public String userProfileName {         get {             return [                     select Profile.Name                     from User                     where Id = :Userinfo.getUserId()                     ].Profile.Name;         }     }     public boolean userAdminTeam {         get {             return [                     select Admin_Team__c                     from User                     where Id = :Userinfo.getUserId()                     ].Admin_Team__c;         }     }      public UserProfileGetter(ApexPages.StandardController unused) {     }   }   And then I use like this:   &lt;apex:tab label="Account Team" name="Account Team" id="Act" rendered="{!userAdminTeam}"&gt;   Thanks all. Hi I don't know how to show that $\langle 2,x \rangle$ is not principal and the definition of a principal ideal is unclear to me. I need help on this, please.   The ring that I am talking about is $\mathbb{Z}[x]$ so $\langle 2,x \rangle$ refers to $2g(x) + xf(x)$ where $g(x)$, $f(x)$ belongs to $\mathbb{Z}[x]$. How to execute a JavaScript function when I have its name as a string How does the PPA fit into the scenario of publishing an application to the Ubuntu Software Center? I've been going through docs for the past couple of hours but I haven't understood what the PPA is? I have a cross-platform Java application that I'd like to publish to the Ubuntu Software Center. My application is open-source and I'm using Github.   Apparently, publishing applications to the store isn't as simple as uploading a deb package - am I right? I need to create an account on Launchpad and put all my code there.  I don't intend to move from Git to Bzr merely for the sake of publishing to the app store but luckily, one is able to set up source-code mirroring from Github to Launchpad.  Since my application is still very premature, it'll have updates fairly often. When I build my application on my machine, do I simply go my Ubuntu App Developer page and upload the new DEB package or do they build my application from source?  What exactly is the PPA for? I don't think I'll need too many of the Launchpad features so I'd like to stick to Github if possible. Is $\le$ defined by $\langle 𝑥_1, 𝑦_1 \rangle \le \langle 𝑥_2, 𝑦_2 \rangle$, if $𝑥_1 \le 𝑥_2 \land𝑦_1 \ge 𝑦_2$ a linear order? I need to tell if the relation $\le$ defined by relationship $\langle 𝑥_1, 𝑦_1 \rangle \le \langle 𝑥_2, 𝑦_2 \rangle$, if $𝑥_1 \le 𝑥_2$ $\land$ $𝑦_1 \ge 𝑦_2$ linear order?   I have already proven that the relation is an order, but I need to decide if it is a linear order and why. Thanks for any help! What does it mean that a package is "set to manually installed?" When packages are already installed and I run an apt-get install &lt;package-name&gt;, sometimes it will print a line &lt;package-name&gt; is set to manually installed.  What does that mean? The infinite product of the sines of all positive integers is zero I tried to prove that $\prod\limits_{k = 1}^\infty  {\sin k = 0}$  Define $$P(m)=\prod\limits_{k = 1}^m  {\left|\sin k\right|}$$  then I take the logarithm of both sides $$\log P(m)=\log \prod\limits_{k = 1}^m  {\left|\sin k\right|}$$  which can be written as $$a_m=\log P(m)=\sum\limits_{k = 1}^m  {\log\,\left|\sin k\right|}$$ for any $k\in\mathbb{N}$ we have $0&lt;\left|\sin k\right|&lt;1$, so all terms of the sum above is strictly negative.  I say that the sequence diverges: $\left\{a_m\right\}_{m\in\mathbb{N}}\to\,-\infty$  This is a delicate point on which I have some doubts.  I know that  $$\int\limits_0^\pi  {\log \left|\sin x\right|{\text{ d}}x}  =  - \pi \log 2$$ and that the function $\log \left|\sin x\right|$ is periodic with period $\pi$,  so the integral from 1 to $\infty$ diverges to $- \infty$. Then the series diverges as well for the integral test for convergence.  As $\log P(m)\to\, -\infty$ we can say that $P(m)\to\,0$ as $m\to\,\infty$  Therefore $$\mathop {\lim }\limits_{m \to \infty } \prod\limits_{k = 1}^m {\left| {\sin k} \right| = \mathop {\lim }\limits_{m \to \infty } \prod\limits_{k = 1}^m {\sin k = \prod\limits_{k = 1}^\infty  {\sin k = } 0} }.\quad \blacksquare$$ How do you sort du output by size? How do you sort du -sh /dir/* by size? I read one site that said use | sort -n but that's obviously not right. Here's an example that is wrong.  [~]# du -sh /var/* | sort -n 0       /var/mail 1.2M    /var/www 1.8M    /var/tmp 1.9G    /var/named 2.9M    /var/run 4.1G    /var/log 8.0K    /var/account 8.0K    /var/crash 8.0K    /var/cvs 8.0K    /var/games 8.0K    /var/local 8.0K    /var/nis 8.0K    /var/opt 8.0K    /var/preserve 8.0K    /var/racoon 12K     /var/aquota.user 12K     /var/portsentry 16K     /var/ftp 16K     /var/quota.user 20K     /var/yp 24K     /var/db 28K     /var/empty 32K     /var/lock 84K     /var/profiles 224M    /var/netenberg 235M    /var/cpanel 245M    /var/cache 620M    /var/lib 748K    /var/spool Before Mavericks, I could use /etc/launchd.conf file to change maximum system resource consumption, for example:  limit maxfiles 16384 unlimited limit maxproc 16384 unlimited   It no longer works in Mavericks.  What is the correct way to do it in the recent version of OS X? I have start learning Selenium WebDriver, but i am confused between Java and Python. Which language should i choose? I do have basic ideas of Core Java but haven't wrote a single piece of code with python.  So please suggest me which is the better language for Selenium WebDriver? Determine the Set of a Sum of Numbers I want to determine the set of natural numbers that can be expressed as the sum of some non-negative number of 3s and 5s.  $$S=\{3k+5j∣k,j∈\mathbb{N}∪\{0\}\}$$  I want to check whether that would be: 0,3, 5, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, and so on.  Meaning that it would include 0, 3, 5, 8. Then from 9 and on, every Natural Number. But how would I explain it as a set? or prove that these are the numbers in the set? Center image element in parent div I was trying to solve the following:  Given: $0 &lt; a &lt; b$ and $n&gt;1$, prove: $$na^{n-1}(b-a) &lt; b^n-a^n &lt; nb^{n-1}(b-a)$$  I managed to get this far using the mean value theorem: $$a^n(b-a)&lt;b^n -a^n&lt;b^n (b-a)$$ Any idea how to continue? How do I bulkify queries for directly or indirectly related records? Regarding the recalc in question rep increase what happens due to the cap? Variablity in cv.glmnet results I am using cv.glmnet to find predictors. The setup I use is as follows:  lassoResults&lt;-cv.glmnet(x=countDiffs,y=responseDiffs,alpha=1,nfolds=cvfold) bestlambda&lt;-lassoResults$lambda.min  results&lt;-predict(lassoResults,s=bestlambda,type="coefficients")  choicePred&lt;-rownames(results)[which(results !=0)]   To make sure the results are reproducible I set.seed(1). The results are highly variable. I ran the exact same code 100 times to see how variable the results were. In the 98/100 runs had one particular predictor always selected (sometimes just on its own); other predictors were selected (co-efficient was non-zero) usually 50/100 times.  So it tells me that each time the cross-validation is running it's going to probably select a different best lambda, because of the initial randomization of the folds matter. Others have seen this problem () but there isn't a suggested solution.  I am thinking that maybe that one which shows up 98/100 is probably pretty highly correlated to all the others? The results do stabilize if I just run LOOCV ($\text{fold-size} = n$), but I am curious why they are so variable when $\text{nfold} &lt; n$. Constraints of relativistic point particle in Hamiltonian mechanics I try to understand constructing of Hamiltonian mechanics with constraints. I decided to start with the simple case: free relativistic particle. I've constructed hamiltonian with constraint:  $$S=-m\int d\tau \sqrt{\dot x_{\nu}\dot x^{\nu}}.$$  Here $\phi=p_{\mu}p^{\mu}-m^2=0$  $-$ .  Then $$H=H_{0}+\lambda \phi=\lambda \phi.$$   So, I want to show that I can obtain from this Hamiltonian the same equation of motion, as obtained from Lagrangian.  But the problem is that I'm not sure what to do with $\lambda=\lambda(q,p)$. I tried the following thing:  $$\dot x_{\mu}=\{x_{\mu},\lambda \phi\}=\{x_{\mu},\lambda p^2\}-m^2\{x_{\mu},\lambda\}=\lambda\{x_{\mu},p^2\}+p^2\{x_{\mu},\lambda\}-m^2\{x_{\mu},\lambda\}$$$$=2\lambda \eta_{\mu b} p^b+p^2\{x_{\mu},\lambda\}-m^2\{x_{\mu},\lambda\}=2\lambda \eta_{\mu b} p^b+p^2\frac{\partial \lambda}{\partial p^{\mu}}-m^2\frac{\partial \lambda}{\partial p^{\mu}},$$  $$\dot \lambda=\{\lambda, \lambda \phi \}=\{\lambda,\lambda p^2\}-m^2\{\lambda,\lambda\}=\lambda\{\lambda,p^2\}+p^2\{\lambda,p^2\}=2\lambda\eta_{ak}p^{a}\frac{\partial \lambda}{\partial x^{k}},$$  $$\dot p_{\mu}=\{p_{\mu},\lambda p^{2}-m^2\lambda \}=p^{2}\{p_{\mu},\lambda\}-m^2\{p_{\mu},\lambda\}=-p^{2}\frac{\partial \lambda}{\partial x^{\mu}}+m^2\frac{\partial \lambda}{\partial x^{\mu}}.$$  If we recall that $p^2-m^2=0$, then we get from the third equation: $\dot p=0$, and from the first: $$\dot x_{\mu}=2\lambda\eta_{ak}p^{a}.$$  So we have    $\dot x_{\mu}=2\lambda\eta_{\mu b}p^{b}.$ $\dot \lambda=2\lambda\eta_{ak}p^{a}\frac{\partial \lambda}{\partial x^{k}}.$ $\dot p=0.$   But I don't know what to do next. Can you help me? I'm trying to set up file transfer via FileZilla and it doesn't seem to matter what details I put in, I always get the following error:     Status: Connection attempt failed with "ECONNREFUSED - Connection   refused by server". Error: Could not connect to server   My Dad has tried on his computer (with FileZilla on Windows) and it connects and everything is fine.. When I try on mine with all the same details that he put in, it comes up with that error.  I read that a firewall might be blocking it so I double checked and my firewall through system preferences and it is off.  Anybody know why this might be???  Additional info:  Macbook Pro running OS X El Capitan version 10.11.6  FileZilla version 3.23.0.2 Is there a word for happiness made possible by a tragic situation? I frequently find myself needing a word to express happiness that acknowledges a tragic or unfortunate circumstance underpinning that happiness. Preferably the word could have both adjective and verb forms. For example, "Gregory [verb-ed | felt adjective about] the balmy February day, suspecting it was yet another consequence of catastrophic and irreversible climate change."  "Bittersweet" is close, but not precisely right. It implies ambivalence rather than genuine gladness. "Schadenfreude" is also in the neighborhood, but I don't want to imply that the tragic part of the situation is what causes the pleasure. (Also, I recognize that the word may not quite exist in English, and I don't know if this is forbidden for this forum, but I'd welcome a loan word or phrase if there's a better one.)