## November 2009

### November 2009

Anyone out there taking the November test?

### Re: November 2009

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

### Re: November 2009

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.

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

### Re: November 2009

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

### Re: November 2009

Thats all of them.Jfekety wrote: I have GR8767, 9367, 9768 and 0568. That's it correct?

### Re: November 2009

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.

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

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

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

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

### Re: November 2009

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!!

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

http://www.mathematicsgre.com/viewtopic.php?f=1&t=290chalmok 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!!

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
**Posts:**30**Joined:**Sat Oct 17, 2009 1:20 pm

### Re: November 2009

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.

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
**Posts:**30**Joined:**Sat Oct 17, 2009 1:20 pm

### Re: November 2009

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

all the best

### Re: November 2009

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