LebGeeks sign up test

Or we could have types of question. Like it'll ask the person what kind of geek they are and questions will be asked towards them. If I'm a hardware guy for example, I could be asked simple stuff like the different types of RAM. If I was a comic book guy, I'd be asked about stuff like Spiderman or something. idk

The idea behind the sign up test is to filter out people willing to do some research and thinking on their own. If you manage to find out tech-related questions that are generalizable (can be randomized) and where you can Google the method but not the exact answer, please send them my way.

If I was a comic book guy, I'd be asked about stuff like Spiderman or something

This is a forum for people who are passionate about technology. Questions about comic books do not apply.

Tarek wroteI saw few posts about the new sign up test and tried it myself ... this test is geared toward programmer , we should have another test for your everyday geek , maybe a small question about hex or binary ...something about hardware perhaps ?

No need for programming knowledge:

To find the sum of multiples of m (not 0) less or equal to n:

Let c = n mod m
S(m,n) = Sum for i=0 -> i=(n-c)/m of m*i
S(m,n) = m * Sum for i=0 -> i=(n-c)/m of i
S(m,n) = m * (n-c)/m * ((n-c)/m + 1) / 2
S(m,n) = (n-c) * (n-c+m) / 2 / m

The final result is S(m1,n) + S(m2,n) - S(m3,n) where m3 is the lcm of m1 and m2.

For more info you can refer to http://lebgeeks.com/forums/viewtopic.php?pid=134266#p134266

I guess programming is easier than that, that's why it exists : D

Tarek wrote

InVader wrote

Tarek wroteI saw few posts about the new sign up test and tried it myself ... this test is geared toward programmer , we should have another test for your everyday geek , maybe a small question about hex or binary ...something about hardware perhaps ?

No need for programming knowledge:

To find the sum of multiples of m (not 0) less or equal to n:

Let c = n mod m
S(m,n) = Sum for i=0 -> i=(n-c)/m of m*i
S(m,n) = m * Sum for i=0 -> i=(n-c)/m of i
S(m,n) = m * (n-c)/m * ((n-c)/m + 1) / 2
S(m,n) = (n-c) * (n-c+m) / 2 / m

The final result is S(m1,n) + S(m2,n) - S(m3,n) where m3 is the lcm of m1 and m2.

For more info you can refer to http://lebgeeks.com/forums/viewtopic.php?pid=134266#p134266

you do realize that being a geek doesn't necessary means to understand math , algorithm and etc ?

My post tackled only the 'geared towards programmers' bit.

But you did raise my curiosity regarding the knowledge of math and/or programming to geeks. Any input on that?

The geeks in question here are rather as per Samer's definition:

samer wroteThis is a forum for people who are passionate about technology. Questions about comic books do not apply.

NuclearVision wroteI guess programming is easier than that, that's why it exists : D

I took the challenge to solve the problem algebraically :)

Tarek wrote

InVader wroteThe geeks in question here are rather as per Samer's definition:

samer wroteThis is a forum for people who are passionate about technology. Questions about comic books do not apply.

how about the other geeks ?

The main page of LebGeeks (www.lebgeeks.com/) clearly states: 'A community for technology geeks in Lebanon since 2004'

I have no idea about non tech geek sites.

as Avo pointed you have hardware geeks , anime and comics geeks , gaming geeks , programming geeks , etc
i ,for example, have zero knowledge in programming

As InVader mentioned, my reply already answers your question:

samer wroteThe idea behind the sign up test is to filter out people willing to do some research and thinking on their own. If you manage to find out tech-related questions that are generalizable (can be randomized) and where you can Google the method but not the exact answer, please send them my way.
This is a forum for people who are passionate about technology. Questions about comic books do not apply.

You don't need to be a programmer to solve the sign-up challenge. You just need to be the kind of person who is willing to invest some time googling around and scratching your head to solve a problem. I know the current challenge may not be the optimal one, that's why I suggested you guys send me better options.

I hope this clarifies the issue. Let's continue this discussion at next week's meeting (as per rule #4) or via PM.