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by raleigh » Sat May 30, 2009 6:24 pm
Are the integers z and f on the right of 0 on the number line?

Restatement: are z and f both positive?

(1) The product of z and f is positive.
Restatement: zf > 0.

Choose z = 1, f = 1. Then zf = 1 > 0, z > 0, f > 0.
Choose z = -1, f = -1. Then zf = 1 > 0, z < 0, f < 0.

Insufficient.

(2) The sum of z and f is positive.
Restatement z + f > 0.

Choose z = 2, f = -1. Then z + f = 1 > 0, z > 0, f < 0.
Choose z = 1, f = 1. Then z + f = 2 > 0, z > 0, f < 0.

Insufficient.

(1)+(2)
(1) zf > 0 tells us that z and f are non-zero. Also, it tells us that z and f are both positive or both negative.

(2) Rules out the case where z and f are both negative since a negative plus a negative is always negative.

So z and f are both positive, and hence to the right of zero on the number line. The answer is C.
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by exhilaration » Sat May 30, 2009 8:21 pm
raleigh wrote:Are the integers z and f on the right of 0 on the number line?

Restatement: are z and f both positive?

(1) The product of z and f is positive.
Restatement: zf > 0.

Choose z = 1, f = 1. Then zf = 1 > 0, z > 0, f > 0.
Choose z = -1, f = -1. Then zf = 1 > 0, z < 0, f < 0.

Insufficient.

(2) The sum of z and f is positive.
Restatement z + f > 0.

Choose z = 2, f = -1. Then z + f = 1 > 0, z > 0, f < 0.
Choose z = 1, f = 1. Then z + f = 2 > 0, z > 0, f < 0.

Insufficient.

(1)+(2)
(1) zf > 0 tells us that z and f are non-zero. Also, it tells us that z and f are both positive or both negative.

(2) Rules out the case where z and f are both negative since a negative plus a negative is always negative.

So z and f are both positive, and hence to the right of zero on the number line. The answer is C.
Thanks raleigh, that makes sense. I sometimes wonder what I was thinking during the exam.
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by raleigh » Sat May 30, 2009 8:56 pm
(2) didn't work out smoothly for me in the second problem. I'll look at it in the morning and see if I can put something together if someone doesn't crack it by then.
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by pratikgandhi » Sun May 31, 2009 2:40 am
Problem 2.

a. n^2/36 is an integer

which means n^2 = 36 * K where K is any integer

n^2 = 3*3*2*2*K

n= 3*2*underroot(K)

so n is divisible by 3

b. 144/n^2 is an integer

which means 144 = n^2 * K where K is any integer

3^2 * 2^4 = n^2 * K

n can be 3 or 2 or 4.. (when K is 16 or 36 or 9 respectively) so we cannot conclusively say whether n would be divisible by 3
trying for a perfect score... 800..
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by raleigh » Sun May 31, 2009 10:53 am
pratikgandhi wrote:Problem 2.

a. n^2/36 is an integer

which means n^2 = 36 * K where K is any integer

n^2 = 3*3*2*2*K

n= 3*2*underroot(K)

so n is divisible by 3
This is incorrect. Just because the number can be written with 3 in the product doesn't make it divisible by 3. Consider a number like 3*sqrt(2). It is NOT divisible by 3 since 3*sqrt(2)/3 = sqrt(2) is NOT an integer.

n is an integer so 3*2*sqrt(k) is an integer since they are equal. We have to prove that k is a perfect square which implies that sqrt(k) is an integer, and we are done.

The problem states that n is a positive integer. n^2/36 is a positive integer means there exists a positive integer k such that n^2/36=k.

Case k is a perfect square:
Then k=m^2 where m is an integer (you can argue that it must be positive, but that won't be important in the argument).

Then n = 2*3*sqrt(k) = 2*3*m so n is divisible by 3.

Case k is not a perfect square:
Assume k is not a perfect square. Then sqrt(k) is irrational. Prove by contradiction. Assume that n is divisible by 3. Then n/3=a where is a positive integer. So n = 3a.

We have two different ways to express the value of n. Set them equal to each other:

3a = 2*3* sqrt(k)
sqrt(k) = 3a/6
sqrt(k) = a/2

But a/2 of a rational number. This contradicts the fact that sqrt(k) is an irrational number (cannot be written as a ratio of integers). So k must be a perfect square.

You definitely shouldn't waste your time writing math proofs on the exam. I don't see a quick solution.
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by raleigh » Sun May 31, 2009 10:59 am
(2) is messier than (1). I'm going to make some coffee and look at this again; there has to be an easier way.
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by aj5105 » Mon Jun 01, 2009 4:53 am
Statement (1)

n ^2 = (36 * 1)^2 or (36 * 2)^2 or (36 * 3)^3

For all, n is divisible by 3.

Sufficient

Statement (2)

144/n^2 = I (I = Integer)

144/I = n^2

Consider I = 144, we get n^2 = 1, n = 1. 1 is not divisible by 3.

Consider I = 4, we get n^2 = 36, n = 6. 6 is divisible by 3.

Insufficient
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by mike22629 » Mon Jun 01, 2009 5:25 am
According to Ron Purewal, always try to disprove DS.

For Second Statement,

144/n^2 = Integer

What if n = sqrt(12)

Then 144/n^2=12, but n is NOT divisble by 3

What is n = 12

Then 144/n^2 =1, and n IS divisible by 3

Hence Statement 2 is insufficient
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by ghacker » Wed Jun 03, 2009 8:23 pm
Question 2

Statement I n^2/36 is an integer that means n^2 will have at least the factors of 36 so n^2 = 2^2*3^2*K for some positive integer K ,
we see that n^2 has 3^2 in it so n must have at least a 3 as a factor

n is divisible by 3 --------------------------- statement I is sufficient

statement II

N^2 must have at least one factor of 144 , ie: 1,2,3,6,16,4,12....

hence if n^2=1 n=1 but n is not divisible by 3 if n^2 = 144 , n=12 and n is divisible by 3 so statement II is insufficient
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