Find the three smallest positive integers $K$ (two digits or greater) with the following properties:      1) $K=\frac{(n)(n+1)}{2}$ for some $n$      2) Each digit of $K$ is the same.   I was able to find the three $K's$ which are $55,66,666$ However, I am looking for a more systematic way. I was able to find the three numbers by testing $n$ I got the following:  For:  $55: n=10 $  $66: n=11 $  $666: n=36$  I am thinking about how can someone solve this problem without a calculator. I believe this problem is written to use a certain idea but I can't seem to find it. Does anyone see it?   As for my research, I've been reading about triangular numbers with the same digits. Someone suggested that I should $(n)$ and $(n+1)$ in prime factorization form.  I noticed that you can't get $6666$  If you try $n=115$, you get $6670$. So $6666$ is not possible.    So basically, I am looking for strategies in how to approach this problem. Any ideas?
For which $x$ exists sum $1 + 2 + 3 + \ldots + n$, where $n &gt; 3$, which has notation $xxx\ldots x$?   In other words, I am looking for a sum of natural numbers which gives a result which has all same digits, e.g. $5555555$ or $22222222222222222$, or whatever else.  How to find this series and how to proof it for all possible series?  I have found the following examples:   $1 + 2 + 3 + \ldots + 10 = 55$ $1 + 2 + 3 + \ldots + 11 = 66$ $1 + 2 + 3 + \ldots + 36 = 666$     And what about next series which ends with (different) repeating $x$? Hot to find and proof it?
What is the idiom, proverb for "Little problems often become big problems if no one takes the initiative to correct them" What is the idiom, proverb for "Little problems often become big problems if no one takes the initiative to correct them"  Which means in an example that If the employees don’t bother to report a malfunctioning machine or a slip-and-fall hazard, serious injuries could occur – injuries that could have been preventable. Ignoring any hazard can be detrimental to one's business and the safety of their employees. Environment variable substitution in sed If I run these commands from a script:  #my.sh PWD=bla sed 's/xxx/'$PWD'/' ... $ ./my.sh xxx bla   it is fine.  But, if I run:  #my.sh sed 's/xxx/'$PWD'/' ... $ ./my.sh $ sed: -e expression #1, char 8: Unknown option to `s'    I read in tutorials that to substitute environment variables from shell you need to stop, and 'out quote' the $varname part so that it is not substituted directly, which is what I did, and which works only if the variable is defined immediately before.  How can I get sed to recognize a $var as an environment variable as it is defined in the shell? Why do damaged hard drives freeze the entire system? Proof Verification: Converse of Intermediate Value Theorem A function $f$ is increasing on $A$ if $f(x)\leq f(y)$ for all $x&lt;y$ in $A$. Show that the Intermediate Value Theorem does have a converse if we assume $f$ is increasing on $[a,b]$.  The converse of IVT along with the additional hypothesis would look like this: Lef $f:[a,b] \rightarrow \mathbb{R}$ be an increasing function on $[a,b]$ which satisfies: if $L$ is a real number such that $f(a)&lt;L&lt;f(b)$ (or $f(a)&gt;L&gt;f(b)$), then there exists a point $c\in (a,b)$ where $f(c)=L.$ If the preceding conditions are met, then $f$ is continuous on $[a,b]$.  The proof (if it's correct) I'm going to show is only for the $c \in (a,b)$, but I think for end points ($a$ or $b$), the method should be similar.  We want to show that given $c \in (a,b)$, for all $\epsilon&gt;0$, there exists a $\delta&gt;0$ such that $|x-c|&lt;\delta \implies |f(x)-f(c)|&lt;\epsilon$.  Since $f$ is increasing, we know that $f(a)\leq f(c)$. If $f(c)-\epsilon&lt;f(a)$, then set $x_1=a$, if $f(c)-\epsilon\geq f(a)$, then we know by hypothesis that a $x_1&lt;c$ exists such that $f(x_1)=f(c)-\epsilon.$ In either case, we have for $x \in (x_1,c]$ $$ f(c)-\epsilon \leq f(x_1) \leq f(x) \leq f(c) .$$  In a similar way, since we know that $f(c)\leq f(b)$, we can deduce that a $x_2&gt;c$ exists such that, for $x \in [c,x_2)$ $$f(c)\leq f(x)\leq f(x_2)\leq f(c)+\epsilon .$$  Pick $\delta = \min \{c-x_1,x_2-c \}$, we then have $$|x-c|&lt;\delta \implies |f(x)-f(c)|&lt;\epsilon .$$  Is this correct? In the proof that I saw, they originally used $\frac{\epsilon}{2}$, instead of just $\epsilon$, but I think this one works just as well, what do you guys think? How to find out client ID of component for ajax update/render? Cannot find component with expression "foo" referenced from "bar" What is the current URL to create a new Google Document please? Since Google last updated their Docs the old URL does not work. Could someone explain this shebang line which uses sh and then does exec perl? This is from . It's about executing a Perl script.     Finally, if you are unfortunate enough to be on an ancient Unix system   that doesn’t support the magic #! line, or if the path to your   interpreter is longer than 32 characters (a built-in limit on many   systems), you may be able to work around it like this:   #!/bin/sh -- # perl, to stop looping eval 'exec /usr/bin/perl -S $0 ${1+"$@"}' if 0;   Can you explain step by step what's going on here? I'm trying to make it work or encompass, but to no avail.  At execution of just the above, I'm getting this:  /bin/sh: 0: Illegal option -- Let G be a group, where $(ab)^3=(a^3)(b^3)$ and $(ab)^5=(a^5)(b^5)$. Prove that $G$ is abelian group.  Thank you in advance. Any help is appreciated. How do I get the path of the directory in which a  script is located, inside that script? I want to use a Bash script as a launcher for another application. I want to change the working directory to the one where the Bash script is located, so I can operate on the files in that directory, like so: $ ./application What are the best ways to improve a site's position in the Google search results? For a given inner product space $V$, I would like to prove the inequality $$\|\mathbf{x}+\mathbf{y}\|^2 + \|\mathbf{x}-\mathbf{y}\|^2 = 2(\|\mathbf{x}\|^2 + \|\mathbf{y}\|^2).$$  We see that $$\|\mathbf{x}+\mathbf{y}\|^2 + \|\mathbf{x}-\mathbf{y}\|^2 = \langle\mathbf{x}+\mathbf{y},\mathbf{x}+\mathbf{y}\rangle + \langle\mathbf{x}-\mathbf{y},\mathbf{x}-\mathbf{y}\rangle,$$  which by linearity $$=\langle\mathbf{x},\mathbf{x}+\mathbf{y}\rangle + \langle\mathbf{y},\mathbf{x}+\mathbf{y}\rangle + \langle\mathbf{x},\mathbf{x}-\mathbf{y}\rangle - \langle\mathbf{y},\mathbf{x}-\mathbf{y}\rangle.$$  I've been trying to figure out how to go about this proof using linearity in the second argument of an inner product, but my textbook does not say that linearity necessarily holds in the second component. If it does, how would I go about the rest of this proof? If not, should I be potentially using some aspect of conjugate symmetry to prove this statement? What will happen when Winter Bash is over? Tomorrow (January 2nd 2019) is the day that Winter Bash will be gone.  So I am thinking: Will there be some sort of a final ranking page anywhere?  It would be nice so I don't need to keep going to  every day to see my rank.  If there isn't going to be anything like that, what should I do to see my final rank on Winter Bash? Suppose that $f$ is analytic on a close curve γ. Prove or disprove $\int_\gamma \overline{f(z)}f'(z)dz$ is purely imagine. Suppose that $f$ is analytic on a close curve γ. Prove or disprove   $$\int_\gamma \overline{f(z)}f'(z)dz$$  is purely imagine.  I know that $f$ is analytic on a close curve, then   $$\int_\gamma f(z) dz=0$$  I tried an example with $\gamma =e^{it}$ with $0\leq t\leq 2\pi$, I often get the real part of the integral equal zero.  I tried let $f=u+iv$ so $f'=u_x+iv_x$. Since $f=u+iv$, $\overline{f}=u-iv$  $$\overline{f(z)}f'(z)=uu_x+vv_x+i(uv_x-vu_x)$$  But this doesn't get me anywhere. Any help would be greatly appreciated. How to add CSS and JS files on node pages from theme's template.php I am trying to add JavaScript and CSS files to one node type page from my template.php file but it is not working.  Here is my template.php code:   &lt;?php // $Id: template.php,v 1.13 2010/12/14 01:04:27 dries Exp $  function mttheme_preprocess_html(&amp;$variables) {   $node = node_load(arg(1));   if ($node-&gt;type = 'sales_team_page') {     drupal_add_css('misc/jquery.fancybox-1.3.4.css');     drupal_add_js('misc/jquery.mousewheel-3.0.4.pack.js',       array('type' =&gt; 'file', 'scope' =&gt; 'header')     );     drupal_add_js('misc/jquery.fancybox-1.3.4.pack.js',       array('type' =&gt; 'file', 'scope' =&gt; 'header')     );     drupal_add_js('misc/jquery.fancybox-calls.js',       array('type' =&gt; 'file', 'scope' =&gt; 'header')     );   } }   Does anyone know why it is not working? About the relation between $\int_{0}^{\infty}\frac{\cosh((\alpha-\beta)x)}{(\cosh(x))^{\alpha+\beta }} \,dx$ and $B(\alpha,\beta)$ How do I get from   $$B(\alpha,\beta)=\int_{0}^{1}{ t^{ \alpha -1}  (1-t)^{ \beta -1} \,dt}$$ to:   $$2^{2- \alpha - \beta } \int_{0}^{ \infty}{ \frac{ \cosh (( \alpha - \beta )x)}{( \cosh(x))^{ \alpha + \beta }} \,dx}$$  ? I have tried different change of variables such as $t=1- \frac{1}{ \cosh (x)}$ and got $\int_{0}^{ \infty}{ \frac{ \sinh (x)  ( \cosh (x)-1)^{ \alpha -1} }{( \cosh(x))^{ \alpha + \beta }} \,dx}$ but without success. I am trying to prove this binomial identity $\sum_{r=0}^n {r {n \choose r}} = n2^{n-1}$ but am not able to think something except induction,which is of-course not necessary (I think) here, so I am inquisitive to prove this in a more general way.  The left side of an identity occurs while solving another problem (concerning binomial theorem) so I am more interested in deriving the right side from the left side, else I have to remember it now onward.  EDIT: I am more interested in an algebraic proof rather than combinatorial argument or something involving calculus (however I liked  and  solution), hence I am removing the combinatorics tag. Why is pruning not needed for random forest trees? Breiman says that the trees are grown with out pruning. Why? I mean to say that there must be a solid reason why the trees in random forest are not pruned. On the other hand it is considered very important to prune a single decision tree to avoid over fitting. Is there some literature available to read for this reason? Of course the trees may not be correlated but there still will be a possibility of over fitting. I was writing some Unit tests last week for a piece of code that generated some SQL statements.  I was trying to figure out a regex to match SELECT, INSERT and UPDATE syntax so I could verify that my methods were generating valid SQL, and after 3-4 hours of searching and messing around with various regex editors I gave up.  I managed to get partial matches but because a section in quotes can contain any characters it quickly expands to match the whole statement.  Any help would be appreciated, I'm not very good with regular expressions but I'd like to learn more about them.  By the way it's C# RegEx that I'm after.  Clarification  I don't want to need access to a database as this is part of a Unit test and I don't wan't to have to maintain a database to test my code. which may live longer than the project. What does it mean to say "a divides b" I am not a number theorist and I am learning about relations.  I encountered a relation that says      $a \leq b$ if $a$ divides $b$   Can someone clarify what it means to a number to divide another number?  Does it mean what I think? $a$ divides $b$ if $a | b \in \mathbb{Z}$?     So given a set $S = \{a, a^2, a^3, \ldots\}$, with relation $a | b \leftrightarrow a \leq b$, does the relation hold   going from left to right or right to left? i.e. $a|a^2, a^2|a^3, \ldots$ How do I reset a lost administrative password? I'm working on a Ubuntu system, and my client has completely forgotten his administrative password. He doesn't even remember entering one; however it is there.  I've tried the suggestions on the website, and I have been unsuccessful in deleting the password so that I can download applets required for running some files. Is there a solution?