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November 2009

Posted: Thu Oct 29, 2009 11:55 am
by Jfekety
Anyone out there taking the November test?

Re: November 2009

Posted: Thu Oct 29, 2009 11:56 am
by sandraizbasa
I will attend it

Re: November 2009

Posted: Thu Oct 29, 2009 12:00 pm
by Jfekety
Cool, it seemed like this forum was dead since October. Are there only four released GREs out there?

Re: November 2009

Posted: Thu Oct 29, 2009 12:16 pm
by origin415
I am also taking it, completely missed the deadline for the october one :/

If you find a fifth, please let us know, but I believe there are only 4 ETS tests.

Re: November 2009

Posted: Thu Oct 29, 2009 12:38 pm
by Jfekety
origin415 wrote:I am also taking it, completely missed the deadline for the october one :/

If you find a fifth, please let us know, but I believe there are only 4 ETS tests.
I have GR8767, 9367, 9768 and 0568. That's it correct?

Re: November 2009

Posted: Thu Oct 29, 2009 3:40 pm
by origin415
Jfekety wrote: I have GR8767, 9367, 9768 and 0568. That's it correct?
Thats all of them.

Re: November 2009

Posted: Thu Oct 29, 2009 4:47 pm
by kosuke
I'll also be taking this thing next week.

Anyone know if the past tests are easier/harder than the "real thing" (the 2009 tests)?

I took the 9768 test last week and it didn't seem as bad as I thought it was going to be.

Re: November 2009

Posted: Fri Oct 30, 2009 8:17 pm
by mrb
Yes.

Re: November 2009

Posted: Sun Nov 01, 2009 7:25 pm
by chalmok
I´m also taking the test for the next week!!

Does somebody knows if there are more tests?? or there are just 4??

Re: November 2009

Posted: Sun Nov 01, 2009 7:29 pm
by chalmok
Can somebody give me a hint with the following problem??

A cyclic group of order 15 has an element x such that the set {x^3, x^5, x^9} has exactly two elements.
The number of elements in the set {x^(13*n) : n is a positive integer} is

A.- 3
B.- 5
C.- 8
D.- 15
E.- infinite

thanks for your help!!

Re: November 2009

Posted: Sun Nov 01, 2009 8:09 pm
by origin415
chalmok wrote:Can somebody give me a hint with the following problem??

A cyclic group of order 15 has an element x such that the set {x^3, x^5, x^9} has exactly two elements.
The number of elements in the set {x^(13*n) : n is a positive integer} is

A.- 3
B.- 5
C.- 8
D.- 15
E.- infinite

thanks for your help!!
http://www.mathematicsgre.com/viewtopic.php?f=1&t=290
Please search if your question has been answered before asking, and if it hasn't start a new thread with the form code and question number.

Re: November 2009

Posted: Mon Nov 02, 2009 1:54 am
by feroz_apple
since the set has only 2 elements...u can deduce that x^3 = e ( as |x| cannot be 2,4 since 2,4 doesnt divide 15)
thus (x^(13n):n} has 3 elements
those are
x^13=x
x^26=x^2
x^39=e
....................and so on.
so we have 3 distinct elements.

Re: November 2009

Posted: Mon Nov 02, 2009 1:56 am
by feroz_apple
and chalmok.......u may ask any question that u are doubtful about...many of us are ready to help rather .........sometimes its too laborious a job to search through volumes of threads ....sometimes the search doesn't work
all the best

Re: November 2009

Posted: Mon Nov 02, 2009 2:28 pm
by chalmok
Thank you very much guys, as you have already noted I´m new in this forum!!