## HELP! GR9768 problems 47,51,54, 61

Forum for the GRE subject test in mathematics.
zhangs2
Posts: 2
Joined: Sun Nov 02, 2008 11:11 pm

### HELP! GR9768 problems 47,51,54, 61

Can any one help me on the following problems!? Thanks alot!!
47. Let x and y be uniformly distributed, independent random variables on [0,1]. The probability that the distance between x and y is less than 1/2 is.
A. 1/4 B. 1/3 C. 12 D. 2/3 E. 3/4 answer: E
I kinda know the answer is E from common sense,, but i dont know how to solve the problem numerically..

51. Let D be the region in the xy-plaine in which the series: sum of ((x+2y)^k) / k from k=1 to infinity converges. Then the interior of D is:

a. an open disk b. the open region bounded by an ellipse
c. the open region bounded by a quadrilateral d. the open region between two parallel lines. e. an open half plane.

I have no idea how to approach this problem..

54. The inside of a certain water tank is a cube measuring 10 feet on each edge and having vertical sides and no top. Let h(t) denote the water level, in feet, above the floor of the tank at time t seconds. Starting at time t = 0, water pours into the tank at a constant rate of 1 cubic foot per second, and simultaneously, water is removed from the tank at a rate of 0.25h(t) cubic feet per second. As t-> infinity, what is the limit of the volume of the water in the tank?
61. what is the greatest integer that divides p^4 - 1 for every prime number p greater than 5?

since I didn't remember learning about the ring of polynomials, also I am very bad at set, subsets these kinds of stuff, I don't understand the problems 57-60 at all.. If any one can explain the basis concepts involved in these problems to me, that will be very thankful! Anyway, I know I should be doing research by myself, it is fine if no one helps, but please do take a look at the previous problems and help me out! Thanks very much!!

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sachem
Posts: 9
Joined: Wed Oct 29, 2008 8:01 pm
47. Let x and y be uniformly distributed, independent random variables on [0,1]. The probability that the distance between x and y is less than 1/2 is.
A. 1/4 B. 1/3 C. 12 D. 2/3 E. 3/4

To approach this problem, I find it is best to draw the region in question - a square in R^2, with X and Y both in [0,1].

The distance between X and Y is less than 1/2. So, |x-y|<1/2.
To draw this region, draw the lines x-y<1/2 (which is the equation assuming x-y is positive) and -(x-y)<1/2 (which is the equation assuming x-y is negative).

Now, if X and Y were random variables with given distributions, you would have to integrate in that region to get the probability. However, since they are uniform, you can just find the area.

51. Let D be the region in the xy-plaine in which the series: sum of ((x+2y)^k) / k from k=1 to infinity converges. Then the interior of D is:

a. an open disk b. the open region bounded by an ellipse
c. the open region bounded by a quadrilateral d. the open region between two parallel lines. e. an open half plane

limit test:
limit of : abs((x+2y)^(k+1)/(k+1) * k/(x+2y)^k )
=abs(x+2y)<1 for this to converge. (and of course, you must test the 'endpoints' - ie abs(x+2y)=1.

However, looking at abs(x+2y)<1, we can break the absolute value down just like before - either x+2y is positive, in which case we have x+2y<1, or x+2y is negative, in which case we have -(x+2y)<1. Sketching this region, you see that it is the region between two parallel lines. You don't even need to check the endpoints, because of the way the answers are given.

54. The inside of a certain water tank is a cube measuring 10 feet on each edge and having vertical sides and no top. Let h(t) denote the water level, in feet, above the floor of the tank at time t seconds. Starting at time t = 0, water pours into the tank at a constant rate of 1 cubic foot per second, and simultaneously, water is removed from the tank at a rate of 0.25h(t) cubic feet per second. As t-> infinity, what is the limit of the volume of the water in the tank?

volume=V=100*h.
V'=100dh/dt=1-.25h
100dh/(1-.25h)=dt
100*-4*ln(1-.25h)=t+c
ln(1-.25h)=-t/400+c'
1-.25h=b*e^(-t/400)
so, as t goes to infinity, 1-.25h=0. Then h=4. So, V=100h=400.

61. what is the greatest integer that divides p^4 - 1 for every prime number p greater than 5?

There are other discussions of this on this board, and it is a pretty tricky question, but here is how the solution goes anyways.

factor as (p^2-1)(p^2+1)=(p-1)(p+1)(p^2+1).
p is odd, so between p-1 and p+1, one of those factors is divisible by 4, and one is divisible by 2. Moreover, p^2 must be odd, so p^2+1 is divisible by 2. =>factor of 16.

You can do a similar argument with the primes 3, and 5, (because you know that (3,p)=1 and (5,p)=1, but it is easiest to use fermat's little theorem, which says that for any prime p, if (a,p)=1, ie. they share no common factors, then a^(p-1)=1(mod p)

then p^2=1(mod 3) so p^2-1 is divisible by 3. =>factor of 3

and, p^4=1(mod 5) so p^4-1 is divisible by 5=> factor of 5

3*5*16 = 240.

57) A Ring has the properties: it is a commutative group under addition, (with identity element called 0), and is closed under multiplication with identity called 1. Other properties have to hold (associativity, distributivity) but these are the most important. Also, if you don't know anything about this, learn that the definition of a field is a ring in which every element is invertible - ie, the multiplicative inverses are in the ring and division can be performed. So, for the problem:

I) All polynomials whose X coefficient is 0. Add any of these two together, and the X coefficient stays zero. Multiply, and the X coefficient similarly stays 0. This is a subring

II)All polynomials whose degree is an even integer, together with the 0 polynomial (note that the question includes "together with the 0 polynomial" because in order for something to be a ring it must have the additive identity, 0).

x^2 is in this ring. So is -x^2+x. However, a ring (or subring) has to be closed under addition. Adding those two, we get x, which is not in the subring. II is not a subring.

III) all polynomials with rational coefficients.

Add or multiply and polynomials with rational coefficients and you only get more rational coefficients. III is a subring.

58)
I)S must be connected, because the function is continuous and defined everywhere. If S was disconnected, the function would have to have a "jump" in it because some intermediate value would be missing.

II) S does not have to be an open subset of R, consider the function f(x)=0. This is closed (ie it contains all of it's limit points, because it has no limit points)

III) if S was not bounded, then f would have to diverge to infinity. However, f is defined everywhere. So, S must be bounded.

59)
A cyclic group is generated by one element, x. So, the cyclic group of order 15 must have elements I, x, x^2,...x^14. This group can be generated by all powers P of x so that (p, 15)=1. This is because if P is some divisor of 15, ie, 5, then we would just get {x^5, x^10, x^15=I}.

we know that {x^3, x^5, x^9} has exactly two elements.
Let's write our elements of the group as I, y,...y^14.

We are given, as choices, that x^13n for any n generates a group (which MUST be a subgroup of the cyclic group of order 15), which has 3, 5, 8, 15 or infinite elements. Rule out infinite - the subgroup must have less or equal elements than the group. Rule out 8 - the order of the subgroup must divide 15. Rule out 15 - because in this case, say, x=y would work, but then x^3, x^5, x^9 would all be the same. Test out 5 - then x could be y^3. But then we have {y^9, y^15 = I, and y^27=y^12} - all are distinct. So, we must have 3 elements in the subgroup, ie x=y^5, and we have {y^15=I, y^25=y^10, and y^45=I} in our subgroup.

Note that this depened on the fact that (13, 15)=1, so if you take any element x in the group, x^13*n will simply be the subgroup generated by x.

60) s=s^2 for all s in ring S. Which must be true?

I) s+s=0. (s+1)^2=s^2+2*s+1 = s+1=s^2+1
s^2+2*s+1=s^2+1 .....so 2*s=s+s=0. I is true

II)(s+t)^2=s+t=s^2+t^2.....this is true
note: careful here, (s+t)^2=s^2+st+ts+t^2....note that st=/=ts necessarily (at least, until we prove that it is commutative). So, we cannot say that (s+t)^2=s^2+2st+t^2=s^2+t^2 (since 2st=0).

III) S is commutative (ie, st=ts). well, (s+t)^2=s^2+t^2=s^2+st+ts+t^2.

So, st+ts=0. st=-ts. However, ts+ts=0 so ts=-ts. Then, st=ts. III is true.

Hope this helped, I'm taking the GRE on saturday so answering questions is helpful for me too. Cheers.

lime
Posts: 129
Joined: Tue Dec 04, 2007 2:11 am
47. Draw the picture and you will see the answer.

51. By the ratio test we have |x+2y|<1 Draw the picture and you will see the answer.

54. dh/dt = 1 - 1/4h
If the limit exists it must be equal to zero, which gives h=4.
Since basement area is 100 square feet, volume would be 4*100 cubic feet.

61.
(p^4 - 1) divisible by 5 (by Fermat's theorem)
(p^2 - 1) divisible by 3 (by Fermat's theorem)
(p-1) divisible by 2 (since p is odd)
(p+1) divisible by 2 (since p is odd)
(p^2+1) divisible by 2 (since p is odd)

zhangs2
Posts: 2
Joined: Sun Nov 02, 2008 11:11 pm
thanks sachem, lime. The explanations are very helpful. i realize that for the problems involving rings and sets, the only way to solve them is to test each choice, which is very time consuming.. I guess I really should give up those problems and spend more time on checking easier ones..
Anyway~~ hope u will do well on the exam! im gonna take the test on 11/8 too.. ..

nlawless
Posts: 1
Joined: Mon Oct 07, 2013 4:45 am

### Re: HELP! GR9768 problems 47,51,54, 61

lime wrote:
61.

(p^4 - 1) divisible by 5 (by Fermat's theorem)

(p^2 - 1) divisible by 3 (by Fermat's theorem)

(p-1) divisible by 2 (since p is odd)

(p+1) divisible by 2 (since p is odd)

(p^2+1) divisible by 2 (since p is odd)
That only shows that 120 divides p^4-1. To show that 240 divides p^4-1, notice that 4 must divide one of p-1 or p+1. Therefore, you get an extra 2 divisor, giving you 240.