Register    Login    Search    Rss Feeds

 Page 1 of 1 [ 7 posts ] 



 
Author Message
 Post subject: integer k ...CAT 6
 Post Posted: Mon Sep 21, 2009 1:20 pm 
Offline


Posts: 5
For positive integer k, is the expression (k + 2)(k2 + 4k + 3) divisible by 4?

(1) k is divisible by 8.
(2) k + 1/3 is an odd integer

NOTE :In the expression K2 is k square.

Solution: The answer posted is A. But, as per my analysis it was D. Can you please share your views about this ??

My Analysis :
B is also right as K+1 is odd, so K is even. The number that leads K to be divisible by 3(K+1) are 2,8 & 14. Which are all divisible by 4 when we input these even integers into the expression.

Waiting for your reply.


Top 
 Post subject: Re: integer k ...CAT 6
 Post Posted: Tue Sep 22, 2009 2:36 pm 
Offline
Students


Posts: 73
(k + 2)(k2 + 4k + 3)

=> k^3 + 4k^2 + 3K + 2K^2 + 8K + 6

A) using A if K is divisible by 8 thn it is alo divisible by 4

however ....+6 is never divisible.... thus answer is always NO >> A / C / D left.


B) the option says K+1/3 is a integer, but i am unable to find any such number.

k = 3 + 1/3
k = 4 + 1/3
....

Thus ...A is the only option left.


Top 
 Post subject: Re: integer k ...CAT 6
 Post Posted: Thu Sep 24, 2009 5:29 am 
Offline
Course Students


Posts: 25
I agree with Answer A ) -->> never divisible by 4.

B) there is no real value of K it seems as K is a positive integer - as per question statement- then you can never have K+1/3 = odd integer --> implies K is a fraction - like 1/3


Top 
 Post subject: Re: integer k ...CAT 6
 Post Posted: Thu Sep 24, 2009 7:11 am 
Offline


Posts: 5
For the second point I am sorry that my question was not specific..It goes like this:

K+1
2) ---- is an odd integer
3

K+1 to be divisible by 3 has to be an odd number, so K is even !!

The numbers that leads to K+1 to be divisible by 3 to be odd are 2,8 & 14.

All went well till I tried 8, by inserting it into the expression..

So, only A is right..

Tx guys.


Top 
 Post subject: Re: integer k ...CAT 6
 Post Posted: Sun Oct 25, 2009 4:25 am 
Offline
ManhattanGMAT Staff


Posts: 823
Let's continue to factor the given expression:
(k+2)(k^2 + 4k + 3)
= (k+2)(k+1)(k+3)
= (k+1)(k+2)(k+3).

In other words, this is the product of three consecutive integers. We can apply consecutive integer, even/odd, and divisibility rules.

There are two ways the answer is "Yes." Either:
- (K+1) and (k+3) are even, OR
- (k+2) is divisible by 4.
The answer is "No" IF
- (k+2) is an even number that is NOT divisible by 4.

Statement (1) says that k is divisible by 8. This means:
(k+1) is odd
(k+2) is even, but is NOT divisible by 4, since k is already divisible by 4
(k+3) is odd.
From statement (1), we know the answer to the question is NO, so statement (1) is sufficient.

From statement (2), (k+1)/3 is an odd integer. Since 3 * (Odd) is still odd, then we know that k + 1 is odd, which makes k even. We know that:
k + 1 is odd
k + 2 is even, but we don't know if it's divisible by 4
k + 3 is odd
So because we don't know if k+2 is divisible by 4, this statement is INSUFFICIENT.


Hope that helps.

_________________
Ben Ku
Instructor
ManhattanGMAT


Top 
 Post subject: Re: integer k ...CAT 6
 Post Posted: Fri Nov 12, 2010 7:03 am 
Offline
Students


Posts: 57
I think there is something wrong with question.. statements contradict each other.. here's how..

When k=2

(K+1)/3 = 3/3 = 1 which is an odd integer so satisfies stat 2

however, statement 1 says k divisible by 8 BUT 2 is NOT divisible by 8.. there s got to be some mistake.. please some prof. check my logic and correct the question or my reasoning, whichever is flawed


Top 
 Post subject: Re: integer k ...CAT 6
 Post Posted: Sun Nov 14, 2010 12:00 am 
Offline
ManhattanGMAT Staff


Posts: 1857
jigar,
The statements actually don't contradict each other. They just give you different sets of numbers.

Statement 1 says k is divisible by 8, so k could be 8, 16, 24, 32, etc.

Statement 2 says that the result of (k+1)/3 is odd, thus k+1 is odd and k is always even, with possible numbers such as 2, 4, 6, 8, 10, and so on.

There is no problem with the statements indicating different sets of numbers as possible values for k. That is completely allowed on the GMAT.

Also, look at it this way--statement 1 indicates values that are a subset of the values of statement 2. If statement two is all positive even integers, then multiples of 8 are contained in that group of numbers. Statement 1 is only indicate a subset of the values in statement 2.

I hope this helps clear things up.

Thank you,

_________________
Jamie Nelson
ManhattanGMAT Instructor


Top 
Display posts from previous:  Sort by  
 
 Page 1 of 1 [ 7 posts ] 





Who is online

Users browsing this forum: No registered users and 1 guest

 
 

 
You cannot post new topics in this forum
You cannot reply to topics in this forum
You cannot edit your posts in this forum
You cannot delete your posts in this forum
You cannot post attachments in this forum

Search for:
Jump to: