It is currently 24 Jan 2018, 01:45

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# For positive integer k, is the expression (k +

 post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
Manager
Joined: 04 Jan 2008
Posts: 118
For positive integer k, is the expression (k + [#permalink]

### Show Tags

16 Sep 2008, 06:38
00:00

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions

### HideShow timer Statistics

This topic is locked. If you want to discuss this question please re-post it in the respective forum.

For positive integer k, is the expression $$(k + 2)(k^2 + 4k + 3)$$ divisible by 4?

(1) k is divisible by 8.

(2) (k + 1)/3 is an odd integer.
Retired Moderator
Joined: 05 Jul 2006
Posts: 1748
Re: Zumit DS 028 [#permalink]

### Show Tags

16 Sep 2008, 06:55
dancinggeometry wrote:
For positive integer k, is the expression $$(k + 2)(k^2 + 4k + 3)$$ divisible by 4?

(1) k is divisible by 8.

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

from one:

(k+2) = even = 2(m+1) where m is devisible by 4..........a

(k^2 + 4k + 3) = even + even+ odd = odd..........b

thus from a & b

th expression boils down to

2*odd*odd thus we can deduct that k is not devisible by 4........ie: 1 is suff

from two

k+1 = odd and thus k is even could be ( check k = 0,2,4) we get multiple answers thus my answer is A
Senior Manager
Joined: 31 Jul 2008
Posts: 289
Re: Zumit DS 028 [#permalink]

### Show Tags

16 Sep 2008, 16:21
A for me as well

2 statement only tells that K is even
Director
Joined: 23 Sep 2007
Posts: 782
Re: Zumit DS 028 [#permalink]

### Show Tags

16 Sep 2008, 19:43
yezz wrote:
dancinggeometry wrote:
For positive integer k, is the expression $$(k + 2)(k^2 + 4k + 3)$$ divisible by 4?

(1) k is divisible by 8.

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

from one:

(k+2) = even = 2(m+1) where m is devisible by 4..........a

(k^2 + 4k + 3) = even + even+ odd = odd..........b

thus from a & b

th expression boils down to

2*odd*odd thus we can deduct that k is not devisible by 4........ie: 1 is suff

from two

k+1 = odd and thus k is even could be ( check k = 0,2,4) we get multiple answers thus my answer is A

Great solution/explanation, but you forgot that k is a positive integer.

D
Intern
Joined: 21 Jun 2008
Posts: 10
Re: Zumit DS 028 [#permalink]

### Show Tags

16 Sep 2008, 22:30
(k+2)(k^2+4k+3) / 4 ***3 is the key...

1) Sufficient because with 8/k, k must be even. therefore, the equation is definitely NOT divisible by 4.

2) Sufficient. Work backwards with (k+1)/3=odd.

--> odd x 3 = odd.
--> k+1 = odd
--> k must be even.
--> if K is even, the equation is definitely NOT divisible by 4, so the final answer is D.
SVP
Joined: 17 Jun 2008
Posts: 1530
Re: Zumit DS 028 [#permalink]

### Show Tags

18 Sep 2008, 00:36
d2touge wrote:

2) Sufficient. Work backwards with (k+1)/3=odd.

--> odd x 3 = odd.
--> k+1 = odd
--> k must be even.
--> if K is even, the equation is definitely NOT divisible by 4, so the final answer is D.

Do not agree with this.

If (k+1)/3 = 1 then original expression will be 3*4*5 and will be divisible by 4.

But, if (k+1)/3 = 3 then original expression will be 9*10*11 and will not be divisible by 4.

Hence, stmt 2 is insufficient and answer should be A.
SVP
Joined: 28 Dec 2005
Posts: 1542
Re: Zumit DS 028 [#permalink]

### Show Tags

19 Sep 2008, 12:58
I agree with A as well ... from stat 2, the only thing I was able to derive was that k is even ...
Re: Zumit DS 028   [#permalink] 19 Sep 2008, 12:58
Display posts from previous: Sort by

# For positive integer k, is the expression (k +

 post reply Question banks Downloads My Bookmarks Reviews Important topics

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.