I computed the 10 first powers of 256 and I noticed that they all end by 6.  256^1 = 256 256^2 = 65536 256^3 = 16777216 256^4 = 4294967296 256^5 = 1099511627776 256^6 = 281474976710656 256^7 = 72057594037927936 256^8 = 18446744073709551616 256^9 = 4722366482869645213696 256^10 = 1208925819614629174706176   My intuition is that it is the same for all powers of 256 but I can't figure out how to prove it, any suggestion?  My attempt was to show that $\forall n$ there is a $k$ such that $2^{8n} = k*10 + 6$, $k$ and $n$ being non null integers. I tried to decompose the right member in powers of $2$ but that leaves me stuck.
We have to prove that  "the product of two numbers ending in 6 also ends with 6" mathematically.  I have no clue how to start. I don't want you to prove it for me! but some hints would be very helpful as I'm totally stuck on how to start this prove.
C# Mono - Low Level Keyboard Hook I am a beginner to python coding in GIS. I want to calculate the total number of counts in raster attribute table. In order to do that:   I would like to build a attribute table of raster data use a loop through attribute table using ARCPY.SearchCursor()   Therefore, I wrote a code to build an attribute table. However, I am getting an error while running the code and not able to figure out the error in the code.  Code:  import arcpy # Create the geoprocessing object  import arcgisscripting gp = arcgisscripting.create()  #To overwrite output arcpy.env.overwriteOutput = True  #Set environment settings gp.workspace = "C:/Subhasis/Test/raster-read"  #Build Raster Attribute Table gp.BuildRasterAttributeTable_management("sti-9", "Overwrite")  Error Message: Traceback (most recent call last):   File "C:/Subhasis/Test/raster-read/test.py", line 15, in &lt;module&gt;     gp.BuildRasterAttributeTable_management("sti-9", "Overwrite") ExecuteError: ERROR 000049: Failed to build attribute table Failed to execute (BuildRasterAttributeTable). Stance on repost question instead of editing to invalidate answers? Why is AES resistant to known-plaintext attacks? I was just testing Eevee and found that you cannot view a glass object through another glass. Is there any workaround? It works fine only If I rotate the glass window or put my camera to the back side of the scene Restrictive vs Non-restrictive Relative Clause Jackknife vs. LOOCV Clone internal HDD to new SSD I recently installed an SSD into my machine. The machine itself is a Lenovo thinkpad W520, and it previously had an internal HDD. I moved the internal HDD into an expansion bay (replacing the CD-ROM), and put the new SSD into the internal bay.   The problem I'm having is that I have my Ubuntu configuration EXACTLY the way I want it - I originally spent many hours configuring it to get it to the way it is now. I'd rather not do this again. But, I'd also like the boot-up gains I would get from the OS being on the SSD.  So, what I'd like to do is clone my Ubuntu partition onto the SSD. The catch is that the standard HDD is significantly bigger than the SSD. And it has a windows partition that I don't need on the SSD (I never use Windows, so if it boots off of the other hard drive, that's fine). The layout of my hard drives are as follows:  /dev/sda (SSD):     Model: ATA M4-CT256M4SSD2 (scsi)     Disk /dev/sda: 256GB     Sector size (logical/physical): 512B/512B     Partition Table: msdos  Number  Start   End    Size   Type     File system  Flags  1      1049kB  147GB  147GB  primary  ext4         boot   /dev/sdb (HDD): Model: ATA ST9500420AS (scsi) Disk /dev/sdb: 500GB Sector size (logical/physical): 512B/512B Partition Table: msdos  Number  Start   End     Size    Type      File system     Flags  1      1049kB  1259MB  1258MB  primary   ntfs            boot  2      1259MB  269GB   268GB   primary   ntfs  4      269GB   483GB   214GB   extended  5      269GB   416GB   147GB   logical   ext4  7      416GB   475GB   58.9GB  logical   linux-swap(v1)  6      475GB   483GB   8470MB  logical  3      483GB   500GB   16.8GB  primary   ntfs   What I've tried so far:  1) Resizing the partitions /dev/sdb5 and /dev/sda1 to be the same size. 2) Booting into Ubuntu 11.04 (from /dev/sdb5) and running dd if=/dev/sdb5 of=/dev/sda1 (of course this causes problems with booting, so I had to reinstall grub.. I can get it to boot, but then I have problems with initrd not finding some files ... presumably it can't load some partitions I think).  Now, I think those two steps are the wrong approach, because it will clone /dev/sdb5 EXACTLY - including references in fstab which point to the wrong hard drive. I'm not sure exactly how to rectify this. I could install Ubuntu 11.04 onto the SSD, then try and copy over all of my configurations, but I'm concerned that I'm going to lose something, or that I'm going to overwrite something like fstab that points back to the original hard drive.   Note that currently, I can still boot from the HDD, so it's not imperative that I get this figured out right away, but I do want it to be exactly how it is right now, so that I can maintain my current level of productivity (it's a work laptop).  Suggestions on how I might be able to overcome this difficulty?   Thanks in advance! Significance of the second focus in elliptical orbits 1.In classical mechanics, using Newton's laws, the ellipticity of orbits is derived. It is also said that the center of mass is at one of the foci.   2.Each body will orbit the center of the mass of the system.  My question is : Are the assumptions in 1 and 2 correct?   Follow up question : Assuming the distance from the centre of the mass to each body remains the same, do we have two bodies orbiting the centre of the mass of the system in an elliptical or circular orbit?   Finally : With elliptical orbits, if the heavier mass is supposed to be in one of the foci, if there is any significance to second focus, what is it? Is it a Lagrange point by any chance or does it have some other mathematical property? Prove by Mathematical Induction: $1(1!) + 2(2!) + \cdot \cdot \cdot +n(n!) = (n+1)!-1$ Prove by Mathematical Induction . . .  $1(1!) + 2(2!) + \cdot \cdot \cdot +n(n!) = (n+1)!-1$  I tried solving it, but I got stuck near the end . . .  a. Basis Step:      $(1)(1!) = (1+1)!-1$      $1 = (2\cdot1)-1$      $1 = 1 \checkmark$   b. Inductive Hypothesis     $1(1!) + 2(2!) + \cdot \cdot \cdot +k(k!) = (k+1)!-1$   Prove k+1 is true.     $1(1!) + 2(2!) + \cdot \cdot \cdot +(k+1)(k+1)! = (k+2)!-1$      $\big[RHS\big]$      $(k+2)!-1 = (k+2)(k+1)k!-1$      $\big[LHS\big]$      $=\underbrace{1(1!) + 2(2!) + \cdot \cdot \cdot + k(k+1)!} + (k+1)(k+1)!$ (Explicit Last Step)      $= \underbrace{(k+1)!-1}+(k+1)(k+1)!$ (Inductive Hypothesis Substitution)      $= (k+1)!-1 + (k+1)(k+1)k!$      $= (k+1)k!-1 + (k+1)^{2}k!$   My [LHS] looks nothing like my [RHS] did I do something wrong?  EDIT:     $ = (k+1)k! + (k+1)^2k! -1 $      $ = (k+1)(k!)(1 + (k+1))-1$      $ = (k+1)(k!)(k+2)-1 = (k+2)(k+1)k!-1$ How can I draw outlines around 3D models? How can I draw outlines around 3D models? I'm referring to something like the effects in a recent Pokemon game, which appear to have a single-pixel outline around them: When and how did "pretty" enter English as an intensifying adverb? Continuous bijection from $(a,b) \to S^1$? Best way to do multi-row insert in Oracle? I'm looking for a good way to perform multi-row inserts into an Oracle 9 database.  The following works in MySQL but doesn't seem to be supported in Oracle.  INSERT INTO TMP_DIM_EXCH_RT  (EXCH_WH_KEY,   EXCH_NAT_KEY,   EXCH_DATE, EXCH_RATE,   FROM_CURCY_CD,   TO_CURCY_CD,   EXCH_EFF_DATE,   EXCH_EFF_END_DATE,   EXCH_LAST_UPDATED_DATE)  VALUES     (1, 1, '28-AUG-2008', 109.49, 'USD', 'JPY', '28-AUG-2008', '28-AUG-2008', '28-AUG-2008'),     (2, 1, '28-AUG-2008', .54, 'USD', 'GBP', '28-AUG-2008', '28-AUG-2008', '28-AUG-2008'),     (3, 1, '28-AUG-2008', 1.05, 'USD', 'CAD', '28-AUG-2008', '28-AUG-2008', '28-AUG-2008'),     (4, 1, '28-AUG-2008', .68, 'USD', 'EUR', '28-AUG-2008', '28-AUG-2008', '28-AUG-2008'),     (5, 1, '28-AUG-2008', 1.16, 'USD', 'AUD', '28-AUG-2008', '28-AUG-2008', '28-AUG-2008'),     (6, 1, '28-AUG-2008', 7.81, 'USD', 'HKD', '28-AUG-2008', '28-AUG-2008', '28-AUG-2008'); For files within a specific folder, I would like vim to never touch "modified" timestamp at all.  The background is that I use  for blogging, which uses plain text files in ~/bloxsom as source of all articles.  The article date (and therefore order of appearence) is based on modification date of the text file. I don't want article pop up like if it's new whenever I just fix a typo. (I do lots of them... :D)  So far, vim changes timestamp and the original stamp is lost forever.  This is OK and I want to keep it that way for most of files on the system.  But I don't want that for the blog files - I can always touch the file if I need.  Any ideas on how to tweak vim into this behavior? Finding the limit of $(1-\cos x)/x^2$ $$\lim _{x \to 0}{1-\cos x\over x^2}={2\sin^2\left(\frac{x}{2}\right)\over x^2}={\frac{2}{x^2}\cdot {\sin^2\left(\frac{x}{2}\right)\over \left(\frac{x}{2}\right)^2}}\cdot\left(\frac{x}{2}\right)^2$$  now $$\lim_{x \to 0}{\sin^2\left(\frac{x}{2}\right)\over \left(\frac{x}{2}\right)^2}=\left({\sin\left(\frac{x}{2}\right)\over \left(\frac{x}{2}\right)}\right)^2=1^2$$  So we have $$\frac{2}{x^2}\cdot \left(\frac{x}{2}\right)^2=\frac{2}{x^2}\cdot \left(\frac{x^2}{4}\right)=\frac{1}{2}$$  Are the moves right? I have a list of structs and I want to change one element. For example :  MyList.Add(new MyStruct("john"); MyList.Add(new MyStruct("peter");   Now I want to change one element:  MyList[1].Name = "bob"   However, whenever I try and do this I get the following error:     Cannot modify the return value of   System.Collections.Generic.List.this[int]‘ because it is not   a variable   If I use a list of classes, the problem doesn't occur.  I guess the answer has to do with structs being a value type.  So, if I have a list of structs should I treat them as read-only? If I need to change elements in a list then I should use classes and not structs? I've got a "diagnostics" page in my ASP.NET application which does things like verify the database connection(s), display the current appSettings and ConnectionStrings, etc.  A section of this page displays the Assembly versions of important types used throughout, but I could not figure out how to effectively show the versions of ALL of the loaded assemblies.  What is the most effective way to figure out all currently referenced and/or loaded Assemblies in a .NET application?  Note: I'm not interested in file-based methods, like iterating through *.dll in a particular directory.  I am interested in what the application is actually using right now. Convergence of $\sum\limits_{n=2}^\infty \frac{1}{n^\alpha \ln^\beta (n)} $ for nonnegative $\alpha$ and $\beta$ Study the convergence of the following series: $$\sum_{n=2}^\infty \frac{1}{n^\alpha \cdot\ln^\beta(n)} \text{ where }\alpha,\beta \geq 0 $$   Applying d'Alembert criterion I have that $$ \lim_{n\to\infty} \frac{n^\alpha  \ln^\beta(n)}{(n+1)^\alpha  \ln^\beta (n+1)} = \lim_{n\to\infty} \left(\frac{n}{n+1}\right)^\alpha\left(\frac{\ln(n)}{\ln (n+1)}\right)^\beta = 1$$ so the nature of the series is inconclusive.  If $\alpha = \beta = 0$, then the series diverges, since $\sum_{n=2}^\infty 1 = \infty$.  Should I study the rest of the cases (i.e. if $\alpha = 0, \beta &gt; 0$ the root test and the ratio test are also inconclusive). What is the best form to study the series?.  Thanks in advance The rule says if the order of two actions/events is clear then the past perfect is not necessary, i.e. you can use past simple. So if the sentence has before or after then the order is clear and therefore the rule applies.  says the first sentence is incorrect. Is he/she right? If so, how do you explain what appears, to me, to be contradictory?     If the Past Perfect is not referring to an action at a specific time,   Past Perfect is not optional. Compare the examples below. Here Past   Perfect is referring to a lack of experience rather than an action at   a specific time. For this reason, Simple Past cannot be used.      Examples:      She never saw a bear before she moved to Alaska. Not Correct      She had never seen a bear before she moved to Alaska. Correct   They say "a lack of experience rather than an action". Saying that just does not solve the problem I have which is the past simple fits the rule of order of events. Not seeing is a negative event with the verb "to see".  I hope the answer won't be this is how we speak or this is how English works or since when languages are logical!