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# 3 GMAT Prep DS - Need help Please

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Intern
Joined: 01 Mar 2006
Posts: 1

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28 May 2007, 18:04
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This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Any help in solving these questions is appreciated ..
Attachments

TocheckGMATPrep1.doc [162.5 KiB]

Last edited by anujkm on 30 May 2007, 19:59, edited 1 time in total.
Director
Joined: 12 Jun 2006
Posts: 532

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28 May 2007, 21:31
h(n) = the prod. of all the even integers from 2 - n.
h(100) = the prod. of all the even integers from 2 - 100; 2(4)6(8) ..... = 2(2*2)(2*3)(2*2*2*) ....
p is defined as the smallest prime factor of h(100)+1. the smalles prime factor of h(100), which is the product of all the even integers b/w 2 and 100, is 2. 2 + 1 =3.
Intern
Joined: 16 May 2007
Posts: 18

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29 May 2007, 10:26
I will skip the first Q as I have no explanation for the answer. It's crazy that you can get that kind of a Q so early:)

For #2:

Since students either like or dislike either food (i will abreviate them "LB" and "BS"), by knowing what they don't like you actually know what they like.

We need the # of students who dislike LB but like BS

You know that 2/3 of the students dislike LB and of those 2/5 like BS (1-3/5). So, in essence, all you need to know is the actual number of students who dislike LB.

(1) Sufficient. Knowing the total number of students in the cafeteria, you can find the # of the students who dislike LB. (120*2/3 = 80)

(2) Sufficient. Number of students who like LB = 1/3 of all students = 40.
You can use that to find the number of students who dislike LB (80).

Intern
Joined: 16 May 2007
Posts: 18

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29 May 2007, 10:32
For Q #3:

The problem tells you that z/y and y/x are integers. I will refer to this as (M)

(1) Sufficient. For zx to be even, at least one of the numbers has to be an even number. However, for (M) to hold true z cannot be an odd number. Which means, it has to be even.

(2) Sufficient. Odd numbers are not divisible by even numbers, so if y is even then z has to be even.

Director
Joined: 03 Sep 2006
Posts: 871

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29 May 2007, 19:30
anujkm wrote:
Any help in solving these questions is appreciated ..

ANswer for the question 1 is straight away result frmo Wilson's Theroem. Please check Wilson's theorem in WIkipedia and remember it.
Director
Joined: 03 Sep 2006
Posts: 871

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29 May 2007, 19:44
anujkm wrote:
Any help in solving these questions is appreciated ..

p divides (p-1)!+1 if p is prime. The converse is also true. For instance, 7 is prime, and 7 divides (6!)+1=721

.
Director
Joined: 03 Sep 2006
Posts: 871

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29 May 2007, 19:49
joroivanov wrote:
I will skip the first Q as I have no explanation for the answer. It's crazy that you can get that kind of a Q so early:)

For #2:

Since students either like or dislike either food (i will abreviate them "LB" and "BS"), by knowing what they don't like you actually know what they like.

We need the # of students who dislike LB but like BS

You know that 2/3 of the students dislike LB and of those 2/5 like BS (1-3/5). So, in essence, all you need to know is the actual number of students who dislike LB.

(1) Sufficient. Knowing the total number of students in the cafeteria, you can find the # of the students who dislike LB. (120*2/3 = 80)

(2) Sufficient. Number of students who like LB = 1/3 of all students = 40.
You can use that to find the number of students who dislike LB (80).

the key as you mentioned is by knowing what they don't like you actually know what they like.
Director
Joined: 03 Sep 2006
Posts: 871

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29 May 2007, 19:54
joroivanov wrote:
For Q #3:

The problem tells you that z/y and y/x are integers. I will refer to this as (M)

(1) Sufficient. For zx to be even, at least one of the numbers has to be an even number. However, for (M) to hold true z cannot be an odd number. Which means, it has to be even.

(2) Sufficient. Odd numbers are not divisible by even numbers, so if y is even then z has to be even.

Is there no possibility that Z might be odd number?? :roll:
Director
Joined: 03 Sep 2006
Posts: 871

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29 May 2007, 19:55
anujkm wrote:
Any help in solving these questions is appreciated ..

You've attached the same file twice. Please remove one of them.
GMAT Club Legend
Joined: 07 Jul 2004
Posts: 5043
Location: Singapore

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29 May 2007, 20:15
Question 1 has been extensively discussed in this forum. You can do a search and find the solutions.
Intern
Joined: 16 May 2007
Posts: 18

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29 May 2007, 20:17
LM - I think not.

if zx is the product of an odd number z and an even number x, then what is stated in the question stem wouldn't hold true (z/y and y/x wouldn't be integers).

If we assume that z is odd, for zx to be even, x has to be even. But then for z/y to be an integer, y has to be odd. However, if y is odd, for y/x to be an integer x has to be odd, which contradicts our first assumption (that x is even).

LM wrote:
joroivanov wrote:
For Q #3:

The problem tells you that z/y and y/x are integers. I will refer to this as (M)

(1) Sufficient. For zx to be even, at least one of the numbers has to be an even number. However, for (M) to hold true z cannot be an odd number. Which means, it has to be even.

(2) Sufficient. Odd numbers are not divisible by even numbers, so if y is even then z has to be even.

Is there no possibility that Z might be odd number??
Intern
Joined: 03 Apr 2006
Posts: 46

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29 May 2007, 20:58
Quote:
Question 1 has been extensively discussed in this forum. You can do a search and find the solutions.

does anyone have a link? I am having trouble finding the past discussion
Intern
Joined: 03 Apr 2006
Posts: 46

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30 May 2007, 07:26
can someone please explain number one or show me the link for the past discussions.
30 May 2007, 07:26
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