# Math GRE Practice Exam

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1. Apr 5, 2008

### PingPong

First off, I'm not sure whether this should be in the homework section since it's not homework, but this seems the best place to get help with it. If it should be moved, please feel free to do so.

These are a few problems I had with one of the GRE practice exams (from one of the Practicing to Take the GRE Mathematics Test books).

20. Suppose that f(1+x)=f(x) for all real x. If f is a polynomial and f(5)=11, then f(15/2) is...
a) -11
b) 0
c) 11
d) 33/2
e) Not uniquely determined

My first reaction was that it couldn't be determined since you can't reduce f(15/2) to f(5). But as I type this, I realize that, if it's a polynomial, it can't keep being 11 at every integer - it must eventually go to infinity or negative infinity - unless it's the constant function f(x)=11. This is the right answer, but is my reasoning correct?

25. Let x and y be positive integers such that 3x+7y is divisible by 11. Which of the following must also be divisible by 11?
a) 4x+6y
b) x+y+5
c) 9x+4y
d) 4x-9y
e) x+y-1

Not sure where to go with this one...:(

48. In the xy-plane, the graph of $x^{\log y}=y^{\log x}$ is:
a) Empty
b) A single point
c) A ray in the open first quadrant
d) a closed curve

My first realization was that there was at least one solution, x=y=1, so it wasn't empty. Then I realized x=y is a solution, so it's not b. I wasn't sure about the others, so I marked c since I had at least a ray, but the answer turns out to be e. Can anybody explain why?

49. If the finite group G contains a subgroup of order 7 but no element other than the identity is its own inverse, then the order of G could be:
a) 27
b) 28
c) 35
d) 37
e) 42

I know it's order must be divisible by 7, but I don't know what the second part about no element being its own inverse implies. The correct answer is c.

EDIT: If there is an element that is its own inverse, then a^2=e. But this implies a subgroup of order 2, so the order of G is not divisible by 2 since there is no a^2=e. The only answer that is divisible by 7 and not by 2 is c, which is the right answer. Is my reasoning sound here?

58. If f(z) an analytic function that maps the entire finite complex plane into the real axis, then the imaginary axis must be mapped to:
a) the entire real axis
b) a point
c) a ray
d) an open finite interval
e) the empty set

My gut reaction was b (which turns out to be the correct answer) but I don't really have proof of it.

Last edited: Apr 5, 2008
2. Apr 6, 2008

### Shooting Star

(The title of the post counts here, it seems. The number of views for this post is very low, despite the problems being tricky in the usual GRE style. Next time, give an awesome name!)

Dear PingPong,

Don't put more than two questions in one post. It becomes a burden for one helper. Divide the questions over a few posts, so that more members can help. One member may not feel comfortable in answering all the questions.
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20. Quite correct.

21. First you have to spot the correct answer, and then justify after coming home.

Notice that x=5 and y=1 gives 3x+7y=22. Putting these in choice (B), you get 11. Just to be sure, try one more combination. See that x=10 and y=2 gives 3x+7y=44, but putting these in (B) does not satisfy. But (D) works for both. It’s probably the correct one.

To prove it now:

If 3x+7y is divisible by 11, so is (3x+7y)+33x
= 36x+7y
= 9(4x-9y)+7y+81y
= 9(4x-9y)+88y. Since 88y is divisible by 11, (4x-9y) must be too.

(Why did I add 33x? Well, I wanted a multiple of 11x which when added to 3x will allow me to get a multiple of 4x, so that I can take out a factor and be left with 4x inside the bracket. 33x is the least of such multiples of 11x.)

48. Taking log of both sides, (log y)(log x) = (log x)(log y). So, it’s an identity, true for all x and y greater than zero, since (log x) is not defined if x is not positive. The open 1st quadrant is the set of all such x,y.

49. Your logic seems to be correct. Notice that I said "seems". I haven't done group related problems for a long time.

58. f(z) = u(x,y) + iv(x,y), where v(x,y) = 0 => $$\partial v/\partial x = \partial v/\partial y = 0.$$

Then, $$\partial u/\partial x = 0 = \partial u/\partial y.$$

If $$\partial u/\partial x = 0$$, then u = g(y) is a function of y only.
If $$\partial u/\partial y = 0$$, then u = h(x) is a function of x only.

This can happen only when u is a real constant C => f(z) = C, which is a point on the real line.

It seems not only the imaginary axis, but whole complex plane is being mapped to a single real number.

Last edited: Apr 6, 2008
3. Aug 7, 2008

### cauchys

To the question number 20, just observe f(x+1)=f(x) and f(5)=11 imply f(n)=5 for n natural number, that means the polynomial g(x)=f(x)-5 has infinite zeros, so g is the zero polynomial,then f(x)=11 for all real x.

Last edited: Aug 7, 2008
4. Aug 7, 2008

5. Aug 7, 2008

### snipez90

First off, why did you revive a few month old thread. Secondly, the first three questions are rather elementary. How much of the GRE is made up of elementary mathematics and calculus? Is it administered by the ETS?

6. Aug 7, 2008

### konthelion

1. There's a big difference between the GRE math section of the General Test and the GRE subject math test. The GRE subject math test consists of material from Calculus(pre-,single,multi), Algebra(abstract,linear), and various other topics(analysis, discrete) whereas the GRE math general test consists of arithmetic, geometry, algebra, and data analysis (probability and graphs).

2. Yes, it's administered by the ETS.