$(5n)!$ has $2014$ trailing zeros. What is $n$?
Possible Duplicate:          I know that I have to find the number of factors of $5$'s, $25$'s, $125$'s etc. in order to do this. But how can you derive such a formula for any number $n$?
Error "Settings to access printer 'printername' are not valid" when printing from Windows service What kinds of compensation can I get from Deutsche Bahn for canceled and late trains? Film where aliens looking like humans are waiting in an airport closed due to snowstorm I saw the movie 6 - 7 years back on cable tv, from the looks of it it seemed to be made in the late 1990's or early 2000's. Unfortunately I could not see the complete movie. The story goes like--   Some people are waiting at a small airport because snowstorm activity has cancelled all flights in and out. There are two children in the group as well (one boy and a girl). Somehow they discover there are aliens looking like humans among them. They decide to check passengers through baggage scanners (checking the skeletal structure via x-ray). They find and kill 2 aliens.   and the big twist:      5. They then realize the two children are also aliens.   any ideas what the name of the movie is ??? I am trying to learn data in higher space into lower space. To have a clue, I'd like to know how to transform the data in the image below into a lower dimension preserving the structure. Hope to hear some explanations and what should I study to learn mapping data to lower dimension? new array.sty and colortbl.sty incompatibility? I just upgraded my texlive distribution and can no longer insert space in table rows using \\[x pt] syntax when colortbl is loaded.  It seems that recent changes to array.sty might be the problem as this is the only package to change recently.  Can anyone suggest a workaround?  MWE  \listfiles \documentclass{article} \usepackage{colortbl} \begin{document} \begin{tabular}{l}   \hline   No space below with \texttt{colortbl.sty} loaded. \\[10mm]   \hline \end{tabular} \end{document}   File versions in use:   article.cls    2014/09/29 v1.4h Standard LaTeX document class   size10.clo    2014/09/29 v1.4h Standard LaTeX file (size option) colortbl.sty    2012/02/13 v1.0a Color table columns (DPC)    array.sty    2018/04/07 v2.4g Tabular extension package (FMi)    color.sty    2016/07/10 v1.1e Standard LaTeX Color (DPC)    color.cfg    2016/01/02 v1.6 sample color configuration   pdftex.def    2018/01/08 v1.0l Graphics/color driver for pdftex I want to install Win Xp on my Ideapad Z 570 laptop but i always get blue screen with message with some drive related problem which asks me to run CHKDSK command. I tried with that but no luck. Is there any way i can install Win Xp with SP 2 ?  Edit: i searched across the sites and came to know there could be some Compatibility problems, and i could fix it by setting HDD mode to IDE; will it solve the problem? Why does ltablex skip a number in captions? The below example presents two tables with an tabularx environment (taken from package ltablex) and a caption. After each tabularx-table, the table counter skips one, as if both the caption and the tabularx environment increase it, while the latter shouldn't.  \documentclass{report} \usepackage{ltablex} \begin{document}  \begin{table} \caption{Caption 1.} % "Table 1" \begin{tabularx}{\linewidth}{|c|} empty table \end{tabularx} \end{table}  \begin{table} \caption{Caption 2.} % "Table 3" \begin{tabularx}{\linewidth}{|c|} empty table \end{tabularx} \end{table}  \end{document}   My reputation doesn't allow me to upload a picture, but the captions are Table 1 and Table 3 respectively. If I call finalize() on an object from my program code, will the JVM still run the method again when the garbage collector processes this object?  This would be an approximate example:  MyObject m = new MyObject();  m.finalize();  m = null;  System.gc()   Would the explicit call to finalize() make the JVM's garbage collector not to run the finalize() method on object m? Rotating image / marker image on Google map V3 How to make traditional output for derivatives What is inertia of a body? The definition of  is  "Inertia is the resistance offered by the body whenever its state of motion is changed."   What is inertia of a body? Is inertia actually a force exerted by the body?  If so, then why is there no application of inertia in numerical problems involving application of force on a body? What is the actual application of inertia in theory? UK coastline shapefile? Let $f_n(x) : \mathbb{R} \rightarrow \mathbb{R}$ be defined as $\dfrac{x}{n}$. Is $f_n$ uniformly convergent?   With "$(a_n)$ is limited and $(b_n)$ converges to 0 $\Rightarrow (a_nb_n)$ converges also to 0" i would guess that $x\cdot \dfrac{1}{n}$ converges uniformly but $x$ is not limited in this case. So does ist converge uniformly or not? Hill Cipher known plaintext attack I know a plaintext - ciphertext couple of length 6 for a hill cipher where its key is a [3x3] matrix.  Based on what I've read and learned, to attack and crack keys of [n x n], if we know a plaintext - ciphertext duo of length $n^2$ then we have our set of n equations with n variables, and this is generally solvable. However in my case there is only length of $6$ instead of $n^2 = 9$ .   My question is, how will I solve this problem and find the key? Or, since this is an obvious homework question, what should be my way of thinking? I cannot think of anything else but matrix multiplications and inverses but they do not help me at all. Show that G is a group, if G is finite, the operation is associative, and cancellation law holds Let $G$ be a non-empty finite set with an associative binary operation so that cancellation law holds, i.e. $ab=ac$ or $ba=ca$ implies $b=c$, for any choices of $a,b,c$ in $G$. Assume that there is an identity element $e$ in $G$. Show that $G$ is a group.  Proof: To show $G$ is a group, conditions must hold. Suppose that $ab=ac$, then $b=eb=(a^{-1}a)b =a^{-1}(ab)=a^{-1}(ac) =(a^{-1}a)c=ec=c$. So left cancellation holds in $G$.  2) Suppose $ba=ca$, then $b=be=b(aa^{-1}) =(ba)a^{-1}=(ca)a^{-1} =c(aa^{-1})=ce=c$. So right cancellation holds in $G$.  3) If $a$ exists in $G$ then $aa^{-1}=a^{-1}a=e$, where $a$ is an inverse of $a^{-1}$. Since inverses are unique, $(a^{-1})^{-1}=a$.  4) Let $x$ be the inverse of $ab$. Then $(ab)x=e$. By associativity we have $a(bx)=aa^{-1}$. Through left cancellation we have  $bx=a^{-1}bx=ea^{-1}=b(b^{-1}a^{-1})$ and $x=b^{-1}a^{-1}$.  Thus $(ab)^{-1}=b^{-1}a^{-1}$. So all conditions hold, $G$ is a group.  Is this proof correct? I know to show $G$ is a group these conditions have to be met. This is all I have to show right? Is there anyway to remove denied permissions from a group for a folder if I dont have access? If light is an electric and (magnetic field), how can it be absorbed? I was wondering how light or any electromagnetic radiation can be "absorbed" if it consists of electric and magnetic fields. For example if there is a charge at point A, and the  light reaches point A, it will cause that charge to move depending on the value of the electric field. However, why wouldn't this electric field continue on and influence others particles everywhere else? For example, if you have a static charge creating an electric field, then if you put a charge at point A it will feel its effects and then if you bring a charge at another location B it will also feel its effects, namely it will not have been absorbed. Please shed some light on this. Thanks  edit: Does this mean the electron that absorbs this light moves in such a way that it creates an electric field of its own canceling the initial one? Or am I way off? Was or is? Confusion My question is for example when closing a call with a customer which is the proper way? "Once again my was (name), have a great day!" or "Once again my is (name), have a great day!"   need this cleared up lol Associates in Integral Domain Let $x$ and $y$ be nonzero elements of an integral domain $D$. Then $x$ and $y$ are associates if and only if $x = yd$ for some unit $d \in D$.    I am done proving the $\Leftarrow$ part. For $\Rightarrow$ what I did was: If $x$ and $y$ are associates then $x \mid y$ and $y \mid x$, so $x \mid y$ implies that $xs = y$ for some $s \in D$ and $y \mid x$ implies that $yt = x$ for some $t \in D$. That is, $yts=y$ which implies $ts=1$. Therefore $s$ is a unit and $t$ is a unit.    Am I on the right track? Can I say that the proof is complete? This is about that old chestnut, $\newcommand{\d}{\mathrm{d}} \int \sin^2 x\,\d x$.   OK, I know that ordinarily you're supposed to use the identity $\sin^2 x = (1 - \cos 2x)/2$ and integrating that is easy. But just for the heck of it, I tried using the $u$-$v$ substitution method (otherwise known as integration by parts).  $$ \int \sin^2 x\,\d x = \int \sin x \sin x\,\d x $$  We can say $u=\sin x$ and $\d u=\cos x\,\d x$ while $\d v = \sin x\,\d x$ and $v = -\cos x$. When we put it all together:   $$\int \sin^2 x\,\d x = u v - \int v\,\d u = -\sin x \cos x - \int -\cos^2 x\,\d x$$  and doing the same routine with $\cos^2 x$ we get $u = \cos x$, $\d u = -\sin x\,\d x$, $\d v = \cos x\,\d x $ and $v = -\sin x$, leading to:  $$\begin{align}   \int \sin^2 x\,\d x = uv - \int v\,\d u     &amp;= -\sin x \cos x - (-\cos x \sin x - \int -\sin^2 x\,\d x) \\     &amp;= -\sin x \cos x + \sin x \cos x - \int \sin^2 x\,\d x \end{align}$$  which eventually works out to  $$2\int \sin^2 x\,\d x = 0$$  So I wanted to get an idea why this didn't work. Maybe it's higher math and the why will be beyond me (I would think that might be the case), or maybe it's one of those proofs that looks absurdly simple when shown that I am just unaware of.