If you sequentially subtract a number from row to row, sometimes excel will get close to, but not 0, even though 0 is the right answer?  For example, if you start out with 1 and subtract 0.05 sequentially from the previous answer, you should eventually get to 0, but excel does not.  Sometimes this poses a problem in doing financial calculations.  Is there a setting that will convert these numbers to exact 0?  I am reluctant to use round() in that it may adversely impact the other numbers that I would like to have the appropriate level of precision.
It's supposed to be zero but for some reason the result produces a peculiar series of digits at the end.  Also, here's the .xlsx of it: http://83.212.111.36/Book1.xlsx
Can the sum of the first $p$ factorials ever be a perfect power for $\ p>3\ $? Has $$\sum_{j=1}^p j!=q^r$$ , where q,p,r are positive integers, and r > 1 , a solution ?   I solved partially, if r is even, then RHS is a perfect square, and there is no doubt in that. Therefore, the left side must be a perfect square as well. But for p>3, the last digit of LHS is 3 which is not the last digit of any square. Hence, p&lt;4, and hence manually checking, only solutions are p =1,3. But i cannot generalize when r is odd.  Any solution to the next part would be helpful! lists Ubuntu 4.10 (Warty Warthog) as the first Ubuntu released by Canonical Ltd.  But I wanted to know whether there ever was an Ubuntu 1.x, as in Windows or Mac OS. If yes, is it available somewhere? Why am I getting objects printed twice? I believe in Windows Vista, we could use linkd command. However, I cannot find this command in Windows 7. I know I can use Windows Explorer and delete the junction, but I wonder if this can be done in Command Prompt.  I can use mklink command to create a junction from a Command Prompt, so, what is the opposite of it (something like linkd)? How can I take a self-portrait without using a remote?  My camera is a Canon Powershot SX210 IS. I do not have a remote, but I do have a tripod. Why is conservation of angular momentum considered a law? We know that angular momentum is defined as the cross product of position and linear momentum. By taking the time derivative, we can deduce that the time rate of change of the angular momentum equals the net torque.  Then why is the conservation of angular momentum considered a law, when we can easily show it is conserved when no net torque is applied, by using Newton's laws?  It truly confuses me. When using Spring Security, what is the proper way to obtain current username (i.e. SecurityContext) information in a bean? I have a Spring MVC web app which uses Spring Security.  I want to know the username of the currently logged in user.  I'm using the code snippet given below .  Is this the accepted way?    I don't like having a call to a static method inside this controller - that defeats the whole purpose of Spring, IMHO.  Is there a way to configure the app to have the current SecurityContext, or current Authentication, injected instead?      @RequestMapping(method = RequestMethod.GET)   public ModelAndView showResults(final HttpServletRequest request...) {     final String currentUser = SecurityContextHolder.getContext().getAuthentication().getName();     ...   } St. Matthew's vs. St. Matthew Cannot invoke method or retrieve property from null object. Object returned by the following call stack is null. \"File\r\n\" This is a really weird situation. A guy contacted me regarding a business I have. He offered far in excess of my stated fees and even told me he could help me financially if I needed it, which I do. He said he was a doctor. Before he even showed up, he gave me his bank account and routing number and full name, which I used to pay some utility bills. All the payments went through. He also sent me $2500 through PayPal, and I received an email that I had received the payment, though it's been 24 hours and it still hasn't shown up in my "recent activity" in my account. But at the time when I met him, I had no reason to doubt the whole thing was real.   So we met, and through conversation he seemed totally legit. Halfway through our session, he abruptly left — he claimed he was being called in to the hospital. He had already received some, but not all, of the services we had agreed to. He said he would come back in a few hours. Oh yeah, and since the PayPal funds hadn't gone through yet, and I needed cash to pay rent, he said his "financial advisor" would be wiring me $800 a couple hours later. Well, I never heard back from him and never got a call about a wire transfer. In retrospect it all seems very obviously a scam, except for the fact that he gave me a bank account number that actually worked as far as paying some bills!  My question is, assuming the PayPal payment never goes through — I never got an email saying that it was cancelled, but it isn't showing up in my payments either. I'm planning on calling PayPal if it hasn't gone through by tomorrow. But assuming that payment doesn't go through, how can I use the bank account info to get the payment he promised me as far as the services I provided? Other than paying all my utilities in advance? You'd think with an account number and routing number and full name, I would be able to withdraw money somehow, but I really have no idea how. And since it isn't under my name, I can't make a payment on my student loans. I wouldn't take more than he owed me, but it would really be nice to at least get that, even if I didn't get the $2500 he had promised me.   You always hear about not letting people see your account number and stuff like that, but it seems impossible to do anything with this info even though I have it! If scamming people with their bank account number is so easy, how do scammers do it? So I'm wondering two things:   What legal way can I take what I am owed from this guy? What illegal way do people use this info if they had it? I don't want to get in trouble, but I'm just curious because you always hear how easy it is.   (Oh, and I never actually saw his ID so I don't know that the name &amp; bank info he gave me was actually his, he could have stolen it from someone, though he really didn't seem like a hacker — he was a middle aged black man who was very smart-sounding, polite, soft spoken, and he seemed to have a lot of knowledge that a real doctor would have. I have a college degree and am very skeptical in general and careful about not being scammed and I've never had an experience even close to this, and the whole time I kept thinking "is this guy for real?" but since the utility stuff had gone through, it seemed legit. And nothing he did sent up a red flag like he might be a criminal or something. But later I googled the name and couldn't find a doctor by that name, at least not in the state he said he lives in. And since he said he would be contacting me last night, and he never did, it seems obvious that he planned the whole thing, but it just makes no sense to me.) Proof that a Combination is an integer From its definition a combination $\binom{n}{k}$, is the number of distinct subsets of size $k$ from a set of $n$ elements.  This is clearly an integer, however I was curious as to why the expression  $$\frac{n!}{k!(n-k)!}$$ always evaluates to an integer.  So far I figured:  $n!$, is clearly divisible by $k!$, and $(n-k)!$, individually, but I could not seem to make the jump to proof that that $n!$ is divisible by their product. I was just thinking about this recently, and I thought of a possible bijection between the natural numbers and the real numbers.  First, take the numbers between zero and one, exclusive.  The following sequence of real numbers is suggested so that we have bijection.    0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9, 0.01, 0.02, ... , 0.09, 0.10, 0.11, ... , 0.99, 0.001, 0.002, ... , 0.999, 0.0001, etc.  Obviously, this includes repeats, but this set is countable.  Therefore, the set of all numbers between zero and one is a subset of the above countable set, and is thus countable.  Then we simply extend this to all real numbers and all the whole numbers themselves, and since the real numbers, as demonstrated above, between any two whole numbers is countable, the real numbers are the union of countably many countable sets, and thus the real numbers are countable.    Please help me with this.  I understand the diagonalization argument by Cantor, but I am curious specifically about this proof which I thought of and its strengths and flaws.  Thanks. I want to prove the assertion:      The unique quadratic subfield of $\mathbb{Q}(\zeta_p)$ is $\mathbb{Q}(\sqrt{p})$ when $p \equiv 1 \pmod{4}$, respectively $\mathbb{Q}(\sqrt{-p})$ when $p \equiv 3 \pmod{4}$.   My first attempt is this. In $\mathbb{Z}[\zeta_p]$, $1-\zeta_p$ is prime and $$ p = \epsilon^{-1} (1-\zeta_p)^{p-1} $$ where $\epsilon$ is a unit.  Since $p$ is an odd prime, $(p-1)/2$ is an integer and  $$ \sqrt{\epsilon p } = (1-\zeta_p)^{(p-1)/2} $$ makes sense and belongs to $\mathbb{Z}[\zeta_p]$.    How do I deal with the $\epsilon$ under the square root?  I guess the condition on the congruence class of $p$ comes from that.  Is this even the right way to proceed?  Uniqueness is not clear to me either.  I thought about looking at the valuation $v_p$ on $\mathbb{Q}$, extending it to two possible quadratic extensions beneath $\mathbb{Q}(\zeta_p)$, then seeing how those have to extend to common valuations on $\mathbb{Q}(\zeta_p)$, but I didn't see how to make it work.  I would appreciate some help. Bluetooth is on but can't find or be found by any other devices.  Output of sudo service bluetooth status:  ● bluetooth.service - Bluetooth service    Loaded: loaded (/lib/systemd/system/bluetooth.service; enabled; vendor preset: enabled)    Active: active (running) since ხუთ 2015-06-04 22:33:18 GET; 13min ago  Main PID: 26678 (bluetoothd)    CGroup: /system.slice/bluetooth.service            └─26678 /usr/sbin/bluetoothd -n  ივნ 04 22:39:14 Rangoo bluetoothd[26678]: Endpoint registered: sender=:1.63 path=/MediaEndpoint/BlueZ4/HFPAG ივნ 04 22:39:14 Rangoo bluetoothd[26678]: Endpoint registered: sender=:1.63 path=/MediaEndpoint/BlueZ4/HFPHS ივნ 04 22:39:14 Rangoo bluetoothd[26678]: Endpoint registered: sender=:1.63 path=/MediaEndpoint/BlueZ4/A2DPSource ივნ 04 22:39:14 Rangoo bluetoothd[26678]: Endpoint registered: sender=:1.63 path=/MediaEndpoint/BlueZ4/A2DPSink ივნ 04 22:39:14 Rangoo bluetoothd[26678]: bluetoothd[26678]: Endpoint registered: sender=:1.63 path=/MediaEndpoint/BlueZ4/HFPAG ივნ 04 22:39:14 Rangoo bluetoothd[26678]: bluetoothd[26678]: Endpoint registered: sender=:1.63 path=/MediaEndpoint/BlueZ4/HFPHS ივნ 04 22:39:14 Rangoo bluetoothd[26678]: bluetoothd[26678]: Endpoint registered: sender=:1.63 path=/MediaEndpoint/BlueZ4/A2DPSource ივნ 04 22:39:14 Rangoo bluetoothd[26678]: bluetoothd[26678]: Endpoint registered: sender=:1.63 path=/MediaEndpoint/BlueZ4/A2DPSink ივნ 04 22:39:14 Rangoo bluetoothd[26678]: Adapter /org/bluez/26678/hci0 has been enabled ივნ 04 22:39:14 Rangoo bluetoothd[26678]: bluetoothd[26678]: Adapter /org/bluez/26678/hci0 has been enabled   Output of lsusb:  Bus 003 Device 005: ID 0a5c:21d7 Broadcom Corp. BCM43142 Bluetooth 4.0   How can I fix it? What is the "continue" keyword and how does it work in Java? I saw this keyword for the first time and I was wondering if someone could explain to me what it does.   What is the continue keyword? How does it work? When is it used? Avoiding errors due to unexpanded asterisk In bash I often use for-loops such as the following  for file in *.type; do    sommecommand "$file";  done;   to perform an operation for all files matching *.type. If no file with this ending is found in the working directories the asterisk is not expanded and usually I will get an error message saying that somecommand didn't find the file. I can immediately think of several ways to avoid this error. But adding a conditional does not seem to be very elegant. Is there a short and clean way to achieve this? custom button visibility list view I have addded a custom button to list views in the Contact object. I have multiple list views but I just need to show the custom button in some of the list views and it should be hidden in the remaining lists views.  Is this possible? Thanks! Service not avaliable Unity3d error in Kubuntu I'm trying to use  in Kubuntu, but I'm getting this error:  Service not avaliable, please try again later.   I've found :  Certain systems may experience "Service unavailable" errors when trying to log in. WORKAROUND: Launch Unity with LD_PRELOAD=/usr/lib/libresolv.so.2 /path/to/Unity (check the path to your local libresolv)   I've checked the path of libresolv.so.2 in my computer. I found it under /lib32/libresolv.so.2 and under /libx32/libresolv.so.2. (Don't know what the difference is).  When I run:  LD_PRELOAD=/lib32/libresolv.so.2 /opt/Unity/Editor/Unity   I get:  ERROR: ld.so: object '/lib32/libresolv.so.2' from LD_PRELOAD cannot be preloaded (wrong ELF class: ELFCLASS32): ignored.   and when I run:   LD_PRELOAD=/libx32/libresolv.so.2 /opt/Unity/Editor/Unity   I get:   ERROR: ld.so: object '/libx32/libresolv.so.2' from LD_PRELOAD cannot be preloaded (cannot open shared object file): ignored.   With both commands I also get this:  /opt/Unity/Editor/Unity: /opt/Unity/Editor/Data/Tools/libgpg-error.so.0: no version information available (required by /lib/x86_64-linux-gnu/libgcrypt.so.20)   Any suggestion? How is stock price determined? Is stock price determined only by the last trade that takes place?  For example, if a stock I own is trading at $3.42, and I place a Sell Limit order for $0.00, will the stock price temporarily dip to $0.00?  Likewise, if I place a Buy Limit order for $100.00 for a stock that is trading at $1.23, will the price spike to $100.00?  Is there some sort of "average price" that is reported on the stock ticker?  Thanks for your help in understanding this! "A/An" preceding a parenthetical statement