Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 06 Feb 2016, 05:49

### 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 +

Author Message
Manager
Joined: 04 Jan 2008
Posts: 119
Followers: 2

Kudos [?]: 39 [0], given: 0

For positive integer k, is the expression (k + [#permalink]  16 Sep 2008, 06:38
00:00

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions
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.
SVP
Joined: 05 Jul 2006
Posts: 1513
Followers: 5

Kudos [?]: 178 [0], given: 39

Re: Zumit DS 028 [#permalink]  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: 306
Followers: 1

Kudos [?]: 32 [0], given: 0

Re: Zumit DS 028 [#permalink]  16 Sep 2008, 16:21
A for me as well

2 statement only tells that K is even
Director
Joined: 23 Sep 2007
Posts: 790
Followers: 5

Kudos [?]: 137 [0], given: 0

Re: Zumit DS 028 [#permalink]  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
Followers: 0

Kudos [?]: 6 [0], given: 0

Re: Zumit DS 028 [#permalink]  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: 1570
Followers: 12

Kudos [?]: 222 [0], given: 0

Re: Zumit DS 028 [#permalink]  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: 1575
Followers: 3

Kudos [?]: 113 [0], given: 2

Re: Zumit DS 028 [#permalink]  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