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My answer is A:
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.

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 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.

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).

Answer D

the key as you mentioned is by knowing what they don't like you actually know what they like.

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.

Answer D

Is there no possibility that Z might be odd number?? :roll:

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.

Answer D

Is there no possibility that Z might be odd number??