text:
# Thread: what version control software do you use?

2. Originally Posted by csn
+1 on subversion. Been using it a while now. So much nicer than CVS.
And another one - I use it standalone (no Apache) with RapidSVN. It's very useful for all those 'why did I make that change?!?!' moments

3. Hi Tim,

You can download lot of about cvs from www.cvshome.org. This site is free and completely for the CVS development only.

I am also in the development phase so won't be able to help u much.

4. Originally Posted by DarkAngelBGE
Yeah, always wanted to use CVs, but had trouble installing it. There were no executables, so I would have go through a 50kb installing ocument, which don't have the time for. :P

Anyone willing to help me installing CVS?

apt-get install cvs

5. I use Subversion (on a linux server) and the Subclipse plugin for Eclipse (WinXP). Works great.

6. There is a Subversion addon that allows you to checkin and checkout from inside Visual Studio .NET, which is my next step.

I've moved from CVSNT and SourceSafe to Subversion and so far, have not regretted it. The only thing that's bothering me is not being able to share files across projects.

7. Thanks vikrant and matrix.

Keep them coming, fellas.

8. *kick*

I am considering using TortoiseSVN. I've read documentation from tigris.org, but I'm not quite sure about some issues. Can someone clear them out for me, before I make some stupid, time-consuming mistakes?

I use one development server: Win2K, Apache, PHP, MySQL. And two clients, a workstation and a laptop for development.
So, do I install Subversion on my server and TortoiseSVN on both clients? How can TortoiseSVN 'connect' to my Subversion server (the documentation doesn't mention this)? Or, should I install TortoiseSVN only on my server?

I have lots of projects where I'm working on, or have been working on. These all fall under one folder ('projects') and are further subdevided. I want to control versions of each of these seperate projects (like: "Project X - update starting-page"). Can I do this in one repository and thus for different (sub-)folders? Could these all be added at once?
Can I backup (and on systemcrash restore) my current files and Subversion's repositories easily?

Thank you very much in advance! It could save me lots of time-waste when someone can answer these relatively simple questions.

9. TortoiseSVN is a client, you are correct, that connects to a subversion server. So you would install subversion on your Win2K server -- good luck with that! -- and you'd install the TortoiseSVN on your laptop & workstation.

Read the Subversion book (see link on http://subversion.tigris.org) to understand how to install the server. Generally the solution is to install it as an Apache [2] module, but you may find it easier to do something else on Windows.

You could certainly use one top-level subversion repository (aka "project") to manage multiple projects. I guess the question you want to ask yourself is why you want them to be part of the same project -- i.e. do you need to create tags (snapshots) accross all projects? If you don't need / want to have revisions tracked accross all of your projects, then you're probably better off making separate subversion repositories for your projects.

As for backup, I'm curious to see what others suggest. The answer should be "yes" -- but there are a number of ways you could automate that, and I don't have any suggestions myself. (Am looking into a similar question for my own repository currently.)

10. Anyone tried CVSNT yet... What are your views on it, is it any good for example? Bad points, etc

11. Umm -- do you mean that Torstoise***CVS*** works great with CVSNT? CVS and Subversion are not compatible, but there is a TortoiseCVS project, which was the original inspiration for TortoiseSVN.

12. Wow, so many Subversion people here. Lately dived into that as well, since we at WIldfire Games were switching from CVS to it regarding our 0 A.D. (the game we are making) codepit.

It seems so much better than CVS.

13. Subversion. I only wish Zend Studio and Dreamweaver had built-in support for it.

In fact, after working with it for a while, I wish everything in Windows was (sub)versioned. Just imagine: turn on versioning on My Documents, and whatever you save there gets versioned. Oh well, maybe after Longhorn

14. My brain and the rename function! I work on my own, and see no need for version control. Someone please try to change my mind.

15. *GONE*

16. Originally Posted by sid egg
My brain and the rename function! I work on my own, and see no need for version control. Someone please try to change my mind.
Programming without version control is like refactoring without unit tests.

If that analogy did not make any sense to you, keep researching all of the terms

17. Originally Posted by sweatje
Programming without version control is like refactoring without unit tests.

If that analogy did not make any sense to you, keep researching all of the terms
I don't use unit testing either I should though.

18. PHP Code:
 $svnUsers++$soloCoders++

19. Hi...

Originally Posted by sid egg
My brain and the rename function! I work on my own, and see no need for version control. Someone please try to change my mind.
Funnily enough I need to undo some changes to some code that I made last week. Unfortunately other developers had been working on the code in the meantime. Here is how I undid the change...
Code:
cvs update -j1.27 -j1.25 classes/htpasswd.php
cvs update -j1.17 -j1.16 unit/htpasswd_test.php
php all_tests.php
cvs commit -m "Reverted screw up with htpasswd file not being found"
Total time about 90 secs and most of that was waiting for tests to run (there are 3000+). I didn't even have to look at the code.

Oh yes, the two hourly Rephlux report told me that the test was failing on the main test sever. You don't think I bother to run these tests myself do you?

yours, Marcus

20. Originally Posted by hlellelid
So you would install subversion on your Win2K server -- good luck with that!
Subversion works just fine on Windows. We run two production servers using it.

Read the Subversion book (see link on http://subversion.tigris.org) to understand how to install the server. Generally the solution is to install it as an Apache [2] module, but you may find it easier to do something else on Windows.
The Apache module also works just fine on Windows. However a simpler approach if you don't need all the Apache goodness is to use the svnserve program that comes with Subversion.

You could certainly use one top-level subversion repository (aka "project") to manage multiple projects. I guess the question you want to ask yourself is why you want them to be part of the same project -- i.e. do you need to create tags (snapshots) accross all projects? If you don't need / want to have revisions tracked accross all of your projects, then you're probably better off making separate subversion repositories for your projects.
We have multiple repositories for multiple projects. Main downside is that out of the box Subversion doesn't give you any way to list all the repositories on one server. So you need to remember what you've called the repository you want to access, or use a third party product. Other than that, separate repositories works well for us, and is practically a requirement if you use svnserve and want to restrict access to different parts of your codebase.

As for backup, I'm curious to see what others suggest. The answer should be "yes" -- but there are a number of ways you could automate that, and I don't have any suggestions myself. (Am looking into a similar question for my own repository currently.)
We just use standard filesystem backup tools. Not sure if this is the best approach but hasn't caused a problem so far

21. Well I am trying to install Subeversion testing it on windows first.. but I can't import my project it says invalid url?

Code:
D:\>svn import d:/project -m "initial import"
svn: Invalid URL 'D:/project'
Can anybody help me set this up I made the 3 directories as described branches, trunk etc

Also I can't find a forum or anything on the original Subversion site so dont know where to get help.

Ok I think i gotthe syntax ight now but it still won't work:S

Code:
C:\Documents and Settings\chris.ICTAPPROACH>svn import c:/project d:/miniserver/
UniServer3_2/diskw/www/test -m "initial import"
svn: Invalid URL 'D:/miniserver/UniServer3_2/diskw/www/test'
pff finally figured it out:

Code:
C:\Documents and Settings\chris.ICTAPPROACH>svn import d:/project file:///d:/min
iserver/UniServer3_2/diskw/www/test/project1 -m "initial import"
Adding         D:\project\tags

22. Hint - use TortoiseSVN

23. Originally Posted by pixelsoul
file:///d:/min
iserver/UniServer3_2/diskw/www/test/project1
What URL did you use when you made your repository? I just used file:///svn/, so for projects I just put them in file:///svn/project1/trunk/. Seems to be working fine so far

Douglas

24. Originally Posted by pixelsoul
Well I am trying to install Subeversion testing it on windows first.. but I can't import my project it says invalid url?

Getting path's and URL's mixed up. URL's always start with a protocol.

25. Originally Posted by hlellelid
. So you would install subversion on your Win2K server -- good luck with that!
Funny, I just installed it on Win2K server, and despite a hiccup with Apache and the Subversion required DLLs (you need to copy them to Apache's module directory), setting it up was really easy. Integrating it into NT's domain authentication using SSPI was also a piece of cake.

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text:
# Matrix Rearrangements - solving a problem with APL.

Given a NxM matrix, one can pick a single element from each row so that the cost of this operation is the smallest. The cost is defined as the sum of every other element left in a row. For each column containing nonzero elements, an element must be picked so that every other element besides the chosen one must originate from different rows. What is the lowest cost for a given matrix?

# Non-alphanumeric C code - a lab notebook

Is it possible to program in C without alphanumeric characters? Is such a subset of C turing-complete? Finally, can we mechanically translate C code to non-alphanumeric C code? Let’s find out!

format: plain


text:
# Multi-Hosts TLS Certificate

It is sometimes convenient to have a domain distributed over two or more machines. This technique, as old as DNS, is interesting to spread the load between multiple hosts, or to provide a bit of high availability. Indeed, if a host becomes inaccessible, at least half of the requests will continue to be successful.

However, since TLS connections have become the norm, and certificates should be renewed automatically, it could be hard to control the validation and the distribution.

I will present you a technique which, with the help of a finely configured web server, allows to get a different certificate on each machine, but usable for the same subdomain.

The trick is, when receiving the verification request, to either serve the file if you have it, or to transmit the request to the next server: it will return the file if it has it, or will return a 404 error.

You have to be careful not to create loops, because you can quickly end up in a situation where no machine has the requested file and therefore calls the next one which does not have it either…

## Use case

This trick is helpfull in a small setup: with more than 3 or 4 servers, it is better to use a deployment strategy between hosts, for example based on ACME client hooks: after validation, the client initiates a connection to the other servers to update their certificate and notify their respective web servers.

I made this installation for my minio servers: each one is accessible through minio.nemunai.re but also individually on minio0.nemunai.re and minio1.nemunai.re. The two servers are in a distinct geographic area, and are not able to execute commands on each other.

So I tried to create two independent certificates, one for each machine:

• the first host must obtain a certificate for minio0.nemunai.re and minio.nemunai.re ;
• the second host must obtain a certificate for minio1.nemunai.re and minio.nemunai.re.

At the end, each of the two servers will be able to answer the requests of minio.nemunai.re, our goal, whatever the status of the other machine.

Let’s Encrypt doesn’t revoke certificates already issued for a domain, so both certificates are valid at the same time for the same domain.

## DNS Configuration

Since we want to keep the architecture simple, we are leaving the load balancing to the DNS: good resolver implementations will do round-robin over our records, ensuring a distribution of clients between each host.

The first step is to create our sub-domains as follows:

minio0.nemunai.re.   3600 IN A 198.51.100.10
minio1.nemunai.re.   3600 IN A 198.51.100.11

This registers a dedicated domain for each host (so we can access it easily when needed). Then we register the common domain with which we will do load-balancing:

minio.nemunai.re.    3600 IN A 198.51.100.10
3600 IN A 198.51.100.11

We do the same for AAAA records, with IPv6 address.

## Reverse Proxy Configuration

We know that when using the ACME http-01 test, we’ll receive a request on minio.nemunai.re/.well-known/acme-challenge. But the problem we have to solve is that when one of the servers will ask for a certificate, the validation request has a 50% chance to arrive on the other host.

This is where we will configure the reverse proxy in order to send the request to the next server if it does not have the answer of the challenge.

On both servers, we’ll handle acme-challenge that way:

  location /.well-known/acme-challenge {
alias /var/lib/acme/webroot;

try_files $uri$uri/ @minio_neighbor;
}

And we also need that the default server handle acme-challenges in the same directory:

server {
listen 80 default;

[...]

location /.well-known/acme-challenge {
alias /var/lib/acme/webroot;
}
}

With the try_files instruction, our reverse proxy will first look for the file in the local directory, then, if it doesn’t exists, it forwards the request to its neighbor server @minio_neighbor.

We declare @minio_neighbor in another block:

  location @storage_neighbor {
proxy_pass http://198.51.100.11;
proxy_set_header X-Real-IP $remote_addr; proxy_set_header X-Forwarded-For$proxy_add_x_forwarded_for;
}

On the other server, we’ll obviously use the address of the other host.

This configuration avoid infinite proxy loop, as if the challenge response is not find localy, we ask the other host, but it’ll not pass through the try_files file as this is the default server that’ll respond. If it’s the other host that have the challenge response, the first server asked will answer it thanks to the proxy, otherwise, the proxify answer will be a 404 error.

And that’s it: we can launch our ACME client on both host, each one can handle the challenge reponse of the other.


text:
BEAST 2 Help Me Choose

# BEAST 2 Help Me Choose Bactrian tree scale operator

## Bactrian Scale Operator

Scale operator that finds scale factor according to a Bactrian distribution (Yang & Rodriguez, 2013), which is a mixture of two Gaussians: p(x) = 1/2N(x;-m,1-m^2) + 1/2N(x;+m,1-m^2) and more efficient than non-Bactrian scale operator.

## kernelDistribution (KernelDistribution)

Provides sample distribution for proposals (optional, default: bactrian) See also kernel distributions.

## tree (Tree)

Beast tree for wich divergence times are scaled.

## scaleFactor (Double)

Magnitude factor used for scaling. Larger values mean bolder proposals, but higher chance of rejection. When optimise is set to true, this value is automatically tuned during MCMC. A starting value of 0.1 seems to be OK in that case.

## rootOnly (Boolean)

Scale root of a tree only, ignored if tree is not specified. For scaling the whole tree, the epoch flex and tree stretch operators are more efficient, so rootOnly=true is recommended.

## optimise (Boolean)

Flag to indicate that the scale factor is automatically changed in order to achieve a good acceptance rate (default true, recommended).

## upper (Double)

Upper Limit of scale factor (optional, default: 0.99999999).

## lower (Double)

Lower limit of scale factor (optional, default: 1.0E-8).

## weight (Double)

Weight with which this operator is selected (required). See also operator weight tuning

## References

Yang Z, Rodríguez CE. Searching for efficient Markov chain Monte Carlo proposal kernels. Proceedings of the National Academy of Sciences. 2013 Nov 26;110(48):19307-12. doi: 10.1073/pnas.1311790110.

## Bayesian evolutionary analysis by sampling trees

Disclaimer: The above is the opinion of the author RB. If you do not agree, or spot a mistake, contact the author, or discuss this in the issues area or raise a new issue. A link will be added from this page to make sure others can find it.

Served through Jekyll, customised theme based on the twentyfourteen wordpress theme.


text:
# Selection rule on angular momentum quantum number for vector operators

According to the Wigner-Eckart theorem, the matrix element of a vector operator in the basis of eigenstates of the square of the magnitude of the angular momentum ($$J^2$$) and the z-component of the angular momentum ($$J_z$$) is proportional to a Clebsch-Gordan coefficient (I am aware that this theorem applies to all spherical tensor operators, but seeing as I'm not very familiar with these other tensor operators, it's best to stick with vector operators, though I'll take an answer discussing more general tensor operators as well). From this, it's concluded that the selection rule on the angular momentum quantum number is

$$\Delta j=0, \pm1 \tag{1}$$

since the relevant Clebsch-Gordan coefficients vanish whenever (1) is violated.

On Wikipedia, the theorem is derived by showing that the matrix elements satisfy the same recursion relations as the Clebsch-Gordan coefficients. The problem is that I don't see how this leads to the selection rules (1). Why can't there also be some other sets of quantities (apart from something directly proportional to the Clebsch-Gordan coefficients) that satisfy the recursion relations, but that don't vanish whenever (1) is violated? The selection rules on $$J_z$$ can be found by directly applying the commutators between $$J_z$$ and the vector operator, without having to worry about the recursion relations. But I've not seen a similar thing being done to $$J^2$$. So how does one arrive at the selection rule (1)?

• Please use MathJax to typeset equations, see physics.stackexchange.com/help/notation for details. Dec 30, 2021 at 20:47
• Should be fixed now. Thanks for pointing out. Dec 30, 2021 at 20:54

The key point is that $$T^{1}_{m_1}\vert \ell_2,m_2\rangle \tag{1}$$ transforms in exactly the same way as \begin{align} \vert 1,m_1\rangle\vert \ell_2,m_2\rangle \, .\tag{2} \end{align} You can see this because \begin{align} L_zT^{1}_{m_1}\vert \ell_2m_2\rangle &= [L_z,T^{1}_{m_1}]\vert \ell 2 m_2\rangle + T^{1}_{m_1}L_z\vert \ell_2 m_2\rangle\, ,\\ &=(m_1+m_2)T^1_{m_1}\vert\ell_2 m_2\rangle \end{align} and likewise the actions \begin{align} L_\pm T^{1}_{m_1}\vert \ell_2 m_2\rangle &=[L_\pm,T^1_{m_1}]\vert \ell_2m_2\rangle + T^{1}_{m_1} L_\pm \vert \ell_2m_2\rangle\, ,\\ &\sim L_\pm \left[\vert 1m_1\rangle \vert \ell_2 m_2\rangle\right] =\left[L_\pm\vert 1m_1\rangle\right]\vert\ell_2 m_2\rangle +\vert 1m_1\rangle\left[L_\pm \vert \ell_2m_2\rangle\right] \end{align} where by $$\sim$$ I mean the same factors with square roots come out, and of course $$[L_\pm,T^{1}_{m_1}]\propto T^1_{m\pm 1}$$.

If you believe that (1) does indeed transform in the same way as (2) does, then it is clear that, if (2) yields states with $$j$$ values ranging from $$\ell_2+1$$ to $$\vert \ell_2-1\vert$$, then the combination in (1) will also yields states with values of $$j$$ in that range. That’s basically the selection rule $$\Delta j=0,\pm 1$$.

Obviously if you do this not with a vector operator but a general tensor $$T^{\ell_1}$$, you will find the range of $$j$$ is just the same as the range in $$\ell_1\otimes\ell_2$$.

• A similar thing was shortly mentioned in Griffiths, but I guess I'm not entirely convinced that just because they transform similarly under a rotation, they have to have the same allowed values of $j$. Had I known why the combination of two angular momenta yield only those specific values of $j$, then perhaps that would have helped, but I imagine deriving the possible values of $j$ would require a much more advanced treatment that is past my level of understanding (right?). Dec 30, 2021 at 22:31
• not really. To obtain the possible values of $j$ just count the values of $m=m_1+m_2$ (with repetitions) and remember that $m_j$ runs from $-j$ to $j$. There is only one $m=\ell_1+\ell_2$ so $j=\ell_1+\ell_2$ must appear once. For this $j$ eliminate $m_j=j,j-1,\ldots,-j$ from your list of $m_1+m_2$ and start again: $m_1+m_2=j-1$ will appear once, etc. until you exhaust your list of $m_1+m_2$. Dec 30, 2021 at 22:48
• Griffiths is not so good for angular momentum coupling. I recommend the appropriate chapter in the (older) text by Gordon Baym. Dec 30, 2021 at 22:49
• Ah, I see. That makes sense. I guess also that since the application of $L_\pm$ and $L_z$ is the same for $T_q^k\left| l_1,m_1 \right>$ as for $\left| k,q \right>\left| l_1,m_1 \right>$, then the application of $L^2$ will also be the same. Then you just need to form the appropiate linear combinations of the states $T_q^k\left| l_1,m_1 \right>$ or $\left| k,q \right>\left| l_1,m_1 \right>$ to find eigenstates of $L^2$, which then will have the same eigenvalues for the two cases (and hence of course the same values of $l$). Dec 31, 2021 at 10:35
• Exactly. You build the combos using the Clebsch’s. Dec 31, 2021 at 10:42


text:
# Math Help - I need help with more factorizaing...

1. ## I need help with more factorizaing...

factorization is not my thing
anywayz here are the two questions

x^2 +3x(x+3y)+2(x+3y)^2

and

3(2x-y)^2 +2y(2x-y)-5y^2

steps would be nice
thank you!

2. Originally Posted by xo-sherry
factorization is not my thing
anywayz here are the two questions

x^2 +3x(x+3y)+2(x+3y)^2

and

3(2x-y)^2 +2y(2x-y)-5y^2

steps would be nice
thank you!
Let $Y = x + 3y$.

So the first becomes

$x^2 + 3xY + 2Y^2$.

$= x^2 + xY + 2xY + 2Y^2$

$= x(x + Y) + 2Y(x + Y)$

$= (x + Y)(x + 2Y)$

$= (x + x + 3y)[x + 2(x + 3y)]$

$= (2x + 3y)(x + 2x + 6y)$

$= (2x + 3y)(3x + 6y)$

$= 3(2x + 3y)(x + y)$.

For the second, let $X = 2x - y$, so it becomes

$3X^2 + 2yX - 5y^2$

$3 \times -5 = -15$ so we need two numbers that add to become $2$ and multiply to become $-15$. They are $5$ and $-3$. So we break up the middle term into $5yX - 3yX$.

$3X^2 + 2yX - 5y^2 = 3X^2 + 5yX - 3yX - 5y^2$

$= X(3X + 5y) - y(3X + 5y)$

$= (3X + 5y)(X - y)$

$= [3(2x - y) + 5y](2x - y - y)$

$= (6x - 3y + 5y)(2x - 2y)$

$= 2(6x + 2y)(x - y)$

$= 4(3x + y)(x - y)$.


text:
# is equal to: Option 1) Option 2) Option 3) Option 4) Option 5)

By applying $(a+b)^{2}=a^{2}+b^{2}+2ab$
$\rightarrow 4\frac{1}{2}$


text:
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If a, b, c are consecutive positive integers and a < b < c, which of

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If a, b, c are consecutive positive integers and a < b < c, which of  [#permalink]

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Updated on: 18 Dec 2017, 23:06
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If a, b, c are consecutive positive integers and a < b < c, which of the following must be true?

I. c - a = 2
II. abc is an even integer
III. (a + b + c)/3 is an integer

A. I
B. II
C. I and II
D. II and III
E. I, II and III

OPEN DISCUSSION OF THIS QUESTION IS HERE: https://gmatclub.com/forum/if-a-b-and-c ... 44907.html

Originally posted by Lolaergasheva on 07 Feb 2011, 23:23.
Last edited by Bunuel on 18 Dec 2017, 23:06, edited 3 times in total.
Renamed the topic, edited the question and added the OA.
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Re: If a, b, c are consecutive positive integers and a < b < c, which of  [#permalink]

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07 Feb 2011, 23:58
1
If a, b, c are consequitive positive integers and a<b<c, which of the following must be true?
1. c-a=2
2. abc is an even integer
3. (a+b+c) /3 is an integer

1.
a=a
b=a+1
c=a+2

c-a=a+2-a=2.
Correct

2. abc
a can be either even or odd
if a=even; abc=even;
if a=odd,b=even; abc=even

Rule:
multiple of any number of integers if atleast once multiplied by an even number will result in even.

even*even=even
even*odd=even

Correct

3. (a+b+c) /3 is an integer
Again;

a=a
b=a+1
c=a+2

a+b+c=(a+a+1+a+2)=3a+3

3a is always divisible by 3
3 is divisible by 3.
Thus "3a+3" is also divisible by 3.

Rule:
if integer x is divisible by n
and
integer y is divisible by n
then; x+y must be divisible by n

Correct.

Ans : "E"
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Re: If a, b, c are consecutive positive integers and a < b < c, which of  [#permalink]

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08 Feb 2011, 01:22
The sum of three consecutive numbers will always be divisible by 3 <<< It's a helpful result. You might want to commit it to your memory
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Re: If a, b, c are consecutive positive integers and a < b < c, which of  [#permalink]

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08 Feb 2011, 03:48
Lolaergasheva wrote:
If a, b, and c are consecutive positive integers and a<b<c, which of the following must be true?
1. c-a=2
2. abc is an even integer
3. (a+b+c) /3 is an integer

A. 1
B. 2
c. 1 and 2
d. 2 and 3
e. 1, 2 and 3

1. c-a=2 --> a, b, c are consequitive positive integers and a<b<c then c=a+2 --> c-a=2. So this statement is always true;

2. abc is an even integer --> out of any 3 consecutive integers at least one must be even thus abc=even. So this statement is also always true;

3. (a+b+c)/3 is an integer --> the sum of odd number of consecutive integers is ALWAYS divisible by that odd number. So this statement is also always true. Or: (a+b+c)/3=(a+a+1+a+2)/3=(3a+3)/3=a+1=integer.

Answer: E.

AmrithS wrote:
The sum of three consecutive numbers will always be divisible by 3 <<< It's a helpful result. You might want to commit it to your memory

Not only that:

• If $$k$$ is odd, the sum of $$k$$ consecutive integers is always divisible by $$k$$. Given $$\{9,10,11\}$$, we have $$k=3$$ consecutive integers. The sum of 9+10+11=30, therefore, is divisible by 3.

• If $$k$$ is even, the sum of $$k$$ consecutive integers is never divisible by $$k$$. Given $$\{9,10,11,12\}$$, we have $$k=4$$ consecutive integers. The sum of 9+10+11+12=42, therefore, is not divisible by 4.

• The product of $$k$$ consecutive integers is always divisible by $$k!$$, so by $$k$$ too. Given $$k=4$$ consecutive integers: $$\{3,4,5,6\}$$. The product of 3*4*5*6 is 360, which is divisible by 4!=24.
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Re: If a, b, c are consecutive positive integers and a < b < c, which of  [#permalink]

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18 Dec 2017, 23:07
Lolaergasheva wrote:
If a, b, c are consecutive positive integers and a < b < c, which of the following must be true?

I. c - a = 2
II. abc is an even integer
III. (a + b + c)/3 is an integer

A. I
B. II
C. I and II
D. II and III
E. I, II and III

OPEN DISCUSSION OF THIS QUESTION IS HERE: https://gmatclub.com/forum/if-a-b-and-c ... 44907.html
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Re: If a, b, c are consecutive positive integers and a < b < c, which of &nbs [#permalink] 18 Dec 2017, 23:07
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If a, b, c are consecutive positive integers and a < b < c, which of

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