Finding the last two digits of $9^{9^{9^{…{^9}}}}$ (nine 9s) I'm continuing on my journey learning about modular arithmetic and got confused with this question:  Find the last two digits of $9^{9^{9^{…{^9}}}}$ (nine 9s). The phi function is supposed to be used in this problem and so far this is what I've got:  $9^{9^{9^{…{^9}}}} ≡ x (\text{mod } 100)$ Where $0 ≤ x ≤ 100$  $9^{9^{9^{…{^9}}}} \text{ (nine 9s) }= 9^a$ In order to know $9^a (\text{mod } 100)$, we need to know $a (\text{mod } \phi(100))$ As $\phi(100)= 40$, we get $a = b (\text{mod } 40)$  $9^{9^{9^{…{^9}}}} \text{ (eight 9s) }= 9^b$ In order to know $9^b (\text{mod } 40)$, we need to know $b (\text{mod } \phi(40))$ As $\phi(40)= 16$, we get $b = c ( \text{mod }16)$  $9^{9^{9^{…{^9}}}}\text{ (seven 9s) }= 9^c $ In order to know $9^c (mod 16)$, we need to know $c (\text{mod } phi(16))$ as $\phi(16)= 8 $ we need to find $c (\text{mod } 8)$  As $9 = 1 (\text{mod } 8)$ $c = 1 (\text{mod } 8)$  I feel like I might have made a mistake somewhere along the way because I'm having a lot of trouble stitching it all back together in order to get a value for the last two digits. Could anyone please help me with this? Thank you!
The last two digits of $9^{9^9}$ I tried to calculate the last two digits of $9^{9^9}$ using Euler's Totient theorem, what I got is that it is same as the last two digits of $9^9$.   How do I proceed further?
Do any clones refuse Order 66? When the clones are issued      telling them to turn on the Jedi,    do any of them refuse to carry out the order? Suppose there are two face down cards each with a positive real number and with one twice the other. Each card has value equal to its number. You are given one of the cards (with value $x$) and after you have seen it, the dealer offers you an opportunity to swap without anyone having looked at the other card.  If you choose to swap, your expected value should be the same, as you still have a $50\%$ chance of getting the higher card and $50\%$ of getting the lower card.  However, the other card has a $50\%$ chance of being $0.5x$ and a $50\%$ chance of being $2x$. If we keep the card, our expected value is $x$, while if we swap it, then our expected value is: $$0.5(0.5x)+0.5(2x)=1.25x$$ so it seems like it is better to swap. Can anyone explain this apparent contradiction? I want to make a full focus on the body, for example: I need to have a clear focus on the each balls in the photo. I used just single point focus.    In this picture I used:   shutter speed: 8 aperture: F/4 ISO: 200 2 flashes, one in the right and one on the top of project camera: nikon d300s lens : 105mm micro 2.8   Thanks Including .eps figure in PDFLatex This question has been asked a lot, but for reason I cannot get it to work. I'd like to include a .eps image in my document - I am using pdflatex in TeXMaker. Below is the input (I am providing the whole preamble incase there is some interference.   \documentclass[11pt]{article} \usepackage{times} \usepackage{graphicx} \usepackage{epstopdf} %converting to PDF \usepackage{amssymb} \usepackage{amsmath} \usepackage{mathtools} \newcommand{\HRule}{\rule{\linewidth}{0.5mm}} \usepackage[parfill]{parskip} \setcounter{secnumdepth}{5} %how deep sections are numbers \usepackage{float} \usepackage[font={sf,footnotesize},labelfont=bf, justification=justified, singlelinecheck=off]{caption} \usepackage[font=footnotesize]{subcaption} \usepackage[scaled]{helvet}  \usepackage{titlesec}  \usepackage[T1]{fontenc} \usepackage[export]{adjustbox} %for large figures \usepackage{upgreek} \usepackage{microtype} \usepackage{setspace} \singlespacing \usepackage[defaultlines=4,all]{nowidow}  \includegraphics[scale=1]{/../model_1_2.eps}    The error I get is:      ! Package pdftex.def Error: File `/model_1_2-eps-converted-to.pdf' not   found.See the pdftex.def package documentation for explanation.Type H    for immediate help.... .../../model_1_2.eps}   I hope someone can help. What approach should I take to establish this logical proof? I need to design a logical math proof:  Write a detailed structured proof to prove that if m and n are integers, then either 4 divides mn or else 4 does not divide n.   Hint: Think about the form of the statement.   I was thinking about first proving that either 4 should divide m or n in order for the first part to be true, then use proof by cases to prove what happened when m was 4, and another case when n was 4...  But I got stuck, what do you recommend:    This is what I got so far: Claim: (m ∈ Z ∧ n ∈ Z) => (4 | mn ˅ 4 ∤ n)  Negated claim: (m ∈ z ∧ n ∈ z) ∧ ( P(n) ∧ ¬P(mn) )  Proof by contradiction  Assume m , n ∈ z:    Assume P(n) ∧ ¬P(mn):       Then ∃ q ∈ R, n = 4q       Let q0 be such that n = 4q0       Then mn = m(4q0)       Then mn = 4(mq0)       Then ∃ q ∈ R, mn = 4q        Then 4 | mn     Then P(n) ∧ P(mn) Then (m , n ∈ z) ∧ P(n) ∧ P(mn))    I know that is wrong, how can I fix it? Cruising the old questions I came across juantheron asking for $\int \frac {\sec x\tan x}{3x+5}\,dx$ He tried using $(3x+5)^{-1}$ for $U$ and $\sec x \tan x$ for $dv$while integrating by parts.  below is his work.  How can I calculate $$ \int {\sec\left(x\right)\tan\left(x\right)  \over 3x + 5}\,{\rm d}x $$  My Try:: $\displaystyle \int \frac{1}{3x+5}\left(\sec x\tan x \right)\,\mathrm dx$  Now Using Integration by Parts::  We get  $$= \frac{1}{3x+5}\sec x +\int  \frac{3}{(3x+5)^2}\sec x\,\mathrm  dx$$  Here he hit his road block.   I tried the opposite tactic   Taking the other approach by parts.   let $$U= \sec x \tan x$$ then$$ du= \tan^2 x \sec x +\sec^3 x$$ and $$dv=(3x+5)^{-1}$$ then $$v=\frac 1 3 \ln(3x+5)$$ Thus $$\int \frac {\sec x \tan x}{3x+5}\,dx= \frac {\ln(3x+5)\sec x \tan x}{3} - \int \frac {\ln(3x+5) [\tan^2 x \sec x +\sec^3 x]}{3} \,dx$$    As you can see I got no further than he did.   So how many times do you have to complete integration by parts to get the integral of the original $\frac {\sec x \tan x}{3x+5} \, dx$ or is there a better way? Is password confirmation necessary on a Change Password dialog? On one hand there is no need to ask for password confirmation because it is annoying. If I made a mistake when confirming my password I'll need retype my very complex password again. There is also a button that can show/hide the password so I don't need retype it in order to confirm it.   However, some users don't use the password show/hide button and the 'forgotten password' option will be pain for them.  Twitter removed the password confirmation from registration but left the dual confirmation in place for changing password. Is this right approach? Taylor expansion for $\arcsin^2{x}$ I stumbled upon this particular expansion that was included in  post.    $$ \displaystyle \arcsin^{2}(x) = \frac{1}{2} \sum_{n=1}^{\infty} \frac{1}{n^{2} \binom{2n}{n}} (2x)^{2n}$$    This caught my eye because I remember trying to derive a Taylor series for $\arcsin^{2}(x)$ a while ago without much success.  Can anyone prove this or point me to a material that would show a proof of this identity?  EDIT :  Feel free to use any mathematical apparatus at hand. I'm not interested in a proof fit for a certain level, nor am I looking for utmost elegance (though that would be lovely). Thousands of .log files in Exchange 2007 Server folder I'm a bit of an Exchange admin noob so I don't know if this is normal, but on our Exchange 2007 server under C:\Program Files\Microsoft\Exchange Server\Mailbox\First Storage Group\ there are 30,000 1MB .log files.  I don't know if those should be there or not, but they keep multiplying like little jackrabbits.  I've so far resisted deleting them as I'm sure something would break, but it's come to the point now where our 80GB disk is completely full up and Exchange has stopped receiving messages.  My best guess is that these are transaction logs and would be useful in rebuilding the database should something go awry, but 30GB of transaction logging seems a bit excessive.  Is there a way to limit it or is that even a good idea?  And of course did I completely miss the point? Is the word "management" singular or plural? Possible Duplicate:          Which one of the following is correct?     Management gets its ideas from its employees.   Management gets their ideas from their employees. Prove that $\lfloor\lfloor x/2 \rfloor / 2 \rfloor = \lfloor x/4 \rfloor$ Why $\dim U+\dim U^\perp=\dim V$? Let $V$ be a vector space. Given $U\le V$, define $U^\perp=\{f\in V^\ast\mid f(u)=0\,,\;\forall u\in U\}$ and given $W\le V^\ast$, define $W^\perp=\{u\in V\mid f(u)=0\;,\;\forall f\in W\}$. This defines a Galois connection between subspaces of $V$ and subspaces of $V^\ast$. If $V$ is finite dimensional, why is $\dim U+\dim U^\perp=\dim V$? Unfortunately, I am not very comfortable with linear algebra. Any hint will be appreciated. I use XeTeX, Version 3.1415926-2.4-0.9998 (TeX Live 2012/W32TeX).  I am using the crop package to add bleed and crop marks to a document. This works fine for pages with white background, but how do I achieve this for pages which have another colour in the bakckground?  I tried the pagecolor package, and see that it somehow extends the pagecolour to around 2 to 3 mm into the bleed but not beyond. How do I make the entire bleed region of the same colour as the page?  \documentclass[12pt]{article} \usepackage[dvips=false,pdftex=false,vtex=false,paperwidth=138mm,paperheight=216mm]{geometry} \usepackage[noaxes,noinfo,cam,dvips,pdftex,center,width=168mm,height=246mm]{crop} \usepackage{xcolor} \usepackage[english]{babel} \usepackage{blindtext} \begin{document} \pagecolor{orange} \blinddocument \end{document} Search across Stack Exchange sites in Android app How to identify predatory publishers/journals So I just started using linux/ubuntu today and I've been getting problems with my wifi, ive tried the solutions on this page:  When im really close to my router, its fine, i can see wifi signals, but when i go just a few steps back it disappears right away. Im using the latest version of ubuntu 17.10  Any suggestions would be nice! :) Find the value of $\sum_0^n \binom{n}{k} (-1)^k \frac{1}{k+1}$ Find the value of $\sum_0^n \binom{n}{k} (-1)^k \frac{1}{k+1}$.  Writing out several terms, I think the answer is $1/(n+1)$, but I'm struggling to prove this. I would greatly appreciate it if someone could show how to prove this. Why can't I define a default constructor for a struct in .NET? The question asks us to classify all distributions in $\mathbb{R}^2$, say $u$, such that $$(xy)u(x,y) = (x^2-y^2)u(x,y)= 0.$$ I am genuinely stuck, any help is extremely appreciated. How to install Wi-Fi driver for Realtek RTL8821CE on Ubuntu 18.04? I'm trying to install drivers for my Wifi on my HP All-in-one desktop. I've been told to show the output of the command "sudo lshw -C network" so here it is:  *-network description: Ethernet interface product: RTL8111/8168/8411 PCI Express Gigabit Ethernet Controller vendor: Realtek Semiconductor Co., Ltd physical id: 0 bus info: pci@0000:01:00.0 logical name: enp1s0 version: 15 serial: 48:ba:4e:5c:06:5f size: 100Mbit/s capacity: 1Gbit/s width: 64 bits clock: 33MHz capabilities: pm msi pciexpress msix bus_master cap_list ethernet physical tp mii 10bt 10bt-fd 100bt 100bt-fd 1000bt 1000bt-fd autonegotiation configuration: autonegotiation=on broadcast=yes driver=r8169 driverversion=2.3LK-NAPI duplex=full firmware=rtl8168h-2_0.0.2 02/26/15 ip=192.168.1.21 latency=0 link=yes multicast=yes port=MII speed=100Mbit/s resources: ire:31 import:e000(size=256) memory: fea04000-fea04fff memory:fea00000-fea03fff  *-network UNCLAIMED description: Network controller product: Realtek Semiconductor Co., Ltd. vendor: Realtek Semiconductor Co., Ltd. physical id: 0 bus info: pci@0000:02:00.0 version: 00 width: 64 bits clock: 33MHz capabilities: pm msi pciexpress cap_list configuration: latency=0 resources: import:d000(size=256) memory:fe900000-fe90ffff   Output from "lspci -nnk | grep -A2 0280":  02:00.0 Network controller [0280]: Realtek Semiconductor Co., Ltd. RTL8821CE 802.11ac PCIe Wireless Network Adapter [10ec:c821] Subsystem: Hewlett-Packard Company RTL8821CE 802.11ac PCIe Wireless Network Adapter [103c:831a] 03:00.0 Unassigned class [ff00]: Realtek Semiconductor Co., Ltd. RTS5229 PCI Express Card Reader [10ec:5229] (rev 01)