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Integers

by chen118 » Mon Jan 16, 2012 5:39 pm
Hi, there are two questions from practise test that I can't solve. Please help.

questions 1: For every positive unteger n, h(n) is defined to be the product of all even integers from 2 to n inclusive. if p is smallest prime facto of h(n), then p is:
A. between 2 & 10
B between 10 and 20
C between 20 and 30
D between 30 and 40
E greater than 40

The answer is A. why?

questions 2:list K consists of 12 consectuive integers, if -4 is the least integer in list K, what's the range of the positive integers in list K?
The answer is 6. Why? I thought it would be 7, counting 12 consecutive integers from -4.

Thank you for your help!
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by neelgandham » Mon Jan 16, 2012 5:56 pm
Question 1. If the question is
For every positive EVEN integer n, the function h(n) is defined to be the product of all the even integers from 2 to n, inclusive. If p is the smallest prime factor of h(100) + 1, then p is

(A) between 2 and 10
(B) between 10 and 20
(C) between 20 and 30
(D) between 30 and 40
(E) greater than 40
Please find my solution here -
https://www.beatthegmat.com/gmat-prep-t97536.html

Question 2: list K consists of 12 consectuive integers, if -4 is the least integer in list K, what's the range of the positive integers in list K?

-4,-3,-2,-1,0,1,2,3,4,5,6,7 is the List. Now the question is what's the range of the positive integers in list K. The list of positive integers is 1,2,3,4,5,6,7 and the range = 7-1(Greatest number - Least number) = 6
Anil Gandham
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by Anurag@Gurome » Mon Jan 16, 2012 7:42 pm
chen118 wrote:Hi, there are two questions from practise test that I can't solve. Please help.

questions 1: For every positive unteger n, h(n) is defined to be the product of all even integers from 2 to n inclusive. if p is smallest prime facto of h(n), then p is:
A. between 2 & 10
B between 10 and 20
C between 20 and 30
D between 30 and 40
E greater than 40

Thank you for your help!
This question has been discussed many times on this forum.


h(100) = 2 * 4 * 6 * ... * 100
= (2 * 1) * (2 * 2) * (2 * 3) * ... * (2 * 50)
= 2^(50) * (1 * 2 * 3 ... * 50)
Then h(100) + 1 = 2^(50) * (1 * 2 * 3 ... * 50) + 1
Now, h(100) + 1 cannot have any prime factors 50 or below, because dividing this value by any of these prime numbers will give a remainder of 1.

The correct answer is E.
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by Anurag@Gurome » Mon Jan 16, 2012 7:48 pm
chen118 wrote:Hi, there are two questions from practise test that I can't solve. Please help.

questions 2:list K consists of 12 consectuive integers, if -4 is the least integer in list K, what's the range of the positive integers in list K?
The answer is 6. Why? I thought it would be 7, counting 12 consecutive integers from -4.

Thank you for your help!
Link K has 12 consecutive integers, in which -4 is the least integer. So, the 12 integers are -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7.
Range of the positive integers in list K = Largest integer - Smallest integer = 7 - 1 = 6
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by chen118 » Mon Jan 16, 2012 8:05 pm
Many thanks to your replies.
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