November 2009

[Jfekety]

Anyone out there taking the November test?

[sandraizbasa]

I will attend it

[Jfekety]

Cool, it seemed like this forum was dead since October. Are there only four released GREs out there?

[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.

[Jfekety]

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?

[origin415]

I have GR8767, 9367, 9768 and 0568. That's it correct?

Thats all of them.

[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.

[mrb]

Yes.

[chalmok]

I´m also taking the test for the next week!!

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

[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!!

[origin415]

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.

[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.

[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

[chalmok]

Thank you very much guys, as you have already noted I´m new in this forum!!