Author 
Message 
TAGS:

Hide Tags

Intern
Joined: 30 Sep 2008
Posts: 11

For positive integer k, is the expression (k + 2)(k^2 + 4k +
[#permalink]
Show Tags
Updated on: 22 Apr 2014, 03:03
Question Stats:
57% (01:24) correct 43% (01:21) wrong based on 870 sessions
HideShow timer Statistics
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
Official Answer and Stats are available only to registered users. Register/ Login.
Originally posted by kishankolli on 22 Sep 2009, 03:51.
Last edited by Bunuel on 22 Apr 2014, 03:03, edited 1 time in total.
Renamed the topic, edited the question and added the OA.




Math Expert
Joined: 02 Sep 2009
Posts: 47898

Re: For positive integer k, is the expression (k + 2)(k2 + 4k +
[#permalink]
Show Tags
22 Apr 2014, 03:10
pretzel wrote: IMO D is the answer.
For second statement, if K + 1/3 or (K+1)/3, K is even. And so, in (K + 2) (K^2 + 4K + 3) = (K+2)(K+3)(K+1) = even x odd x odd
Am I correct? No, the correct answer is A. For positive integer k, is the expression (k + 2)(k^2 + 4k + 3) divisible by 4?\((k + 2)(k^2 + 4k + 3)=(k+1)(k+2)(k+3)\), so the expression is the product of three consecutive integers. (1) k is divisible by 8 > \(k=8n=even\) > \((k+1)(k+2)(k+3)=odd*even*odd\). Now, \(k+2=8n+2\), though even, is not a multiple of 4 (it's 2 greater than a multiple of 8), therefore the expression is not divisible by 4. Sufficient. (2) (k + 1)/3 is an odd integer > \(k+1=3*odd=odd\) > \(k=even\) > \((k+1)(k+2)(k+3)=odd*even*odd\). Now, \(k+2=even\) may or may not be divisible by 8, therefore the expression may or may not be divisible by 8. For example, consider \(k=2\) and \(k=6\). Not sufficient. Answer: A. Hope it's clear.
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics




Intern
Joined: 18 Aug 2009
Posts: 12

Re: integer k ... DS
[#permalink]
Show Tags
22 Sep 2009, 08:31
Start off by factoring the original equation in the stem where (k^2+4k+3)=(k+3)(k+1) So, (k+2(k^2+4k+3)=(k+1)(k+2)(k+3) where (k+1)(k+2)(k+3) are essentially consecutive integers.
As consecutive integers, (k+1)(k+2)(k+3) can be (odd)(even)(odd) in which case k has to be even. It can also be (even)(odd)(even) in which case k has to be odd. If k is odd,(k+1)(k+2)(k+3) will definitely be divisible by 4 as the product terms carry at least two 2s. This is because there are two even numbers in the product terms and they will each carry at least one 2 in their prime factorization and 4=(2)(2).
If k is even however,(k+1)(k+2)(k+3) may or may not be divisible by 4. It depends if the middle term carries two 2s, (i.e. is a multiple of 4) or not.(k+1) and (k+3) will definitely not carry a 2 in this case as they are odd numbers. For example, if k=2,(k+1)(k+2)(k+3) can be (3)(4)(5) which is divisible by 4 as the middle term, 4, is a multiple of 4. The same logic applies for k=6. If k=8 however,(k+1)(k+2)(k+3) = (9)(10)(11) which is NOT divisible by 4 as the middle term is 10=5x2 and there is only one 2 in its prime factorization.
1) All multiples of 8 are even. Hence, k is even. Possible values of (k+1)(k+2)(k+3) are: If k=8, (k+1)(k+2)(k+3) = (9)(10)(11) If k =16, (k+1)(k+2)(k+3) =(17)(18)(19) If k=24,(k+1)(k+2)(k+3) = (25)(26)(27) If k=32, (k+1)(k+2)(k+3) =(33)(34)(25)
Looking at the middle, or the only even term, of each series of consecutive integers, 10, 18 or 26, we see that in their prime factorization, they only contain one 2. Hence, the entire product of 3 consecutive integers will NOT be divisible by 4 if k is a multiple of 8.
Logically, this is also because multiples of 4 which are 8 or greater include  8,12,16,20, 24 Multiples of 8 include 8, 16, 24. The equation given in the stem asks for the product of 3 consecutive integers. The difference of multiples of 4 and 8 are either 0 or a multiple of 8 apart. Never 3 apart. Hence, the middle number of (k+1)(k+2)(k+3) when k is a multiple of 8 will never contain a multiple of 4. Statement A is sufficient.
2) I assume the statement means (k+1)/3 is odd. So, k+1=3(odd number) k+1=odd k=odd1 k = even
For reasons explained in the beginning, if k is even,(k+1)(k+2)(k+3) may or may not be divisible by 4.
So, Statement B is insufficient.
Is the answer A?




Manager
Joined: 11 Sep 2009
Posts: 129

Re: integer k ... DS
[#permalink]
Show Tags
22 Sep 2009, 17:03
I agree that the correct answer is A.
Is (k + 2)(k2 + 4k + 3) divisible by 4?
\((k + 2)(k^2 + 4k + 3)\) \(= (k + 1)(k + 2)(k + 3)\)
Now to simplify matters, one of the terms NEEDS to be divisible by 4 in order for the whole expression to be divisible by 4. Why?
In order for an integer to be divisible by 4, its product must contain two factors of 2. This would appear to be satisfied by two even numbers. In the instance that we had two even numbers (only possible in this case if k+1 and k+3 are even, then we would already have a number divisible by 4 since those numbers are two consecutive even numbers.
Statement 1: k is divisible by 8:
If k is divisible by 8, then k = 8n, where n is a positive integer. This immediately makes k+1 and k+3 odd numbers, which are not divisible by 4. Now we test the (k+2) term:
\(k+2 = 8n + 2 = 4(2n) + 2\)
As a result, if k is divisible by 8, the expression when divided by 4 will have a remainder of 2. Therefore we can can conclude it IS NOT divisible by 4, and Statement 1 is sufficient.
Statement 2: (k + 1)/3 is an odd integer:
We can represent an odd integer as 2n + 1, where n is an integer. Therefore, \((k+1)/3 = 2n + 1\) \(k = 6n + 2\)
Subbing this into the original equation:
\((k+1)(k+2)(k+3)\) \(= (6n + 3)(6n + 4)(6n + 5)\)
Since 6n + 3 and 6n + 5 both represent odd numbers, if 6n + 4 is divisible by 4, the expression is divisible by 4. Since we cannot determine this (if n = 1, it is not divisible by 4, if n = 2, it is divisible by 4, etc.), Statement B is insufficient.
Therefore, the correct answer is A.



Intern
Joined: 30 Sep 2008
Posts: 11

Re: integer k ... DS
[#permalink]
Show Tags
24 Sep 2009, 04:17
Yes..Its A... Its a farily simple answer...But I guess due t the timeing I got it messed it up..
My Analysis :
1>Since its divisible by 8, its divisible by 4...SO TRUE
2.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..I just tried inserting 2...and since I got a positive answer I did not try for 8.
So, only A is right..
Tx guys.



Senior Manager
Joined: 22 Dec 2009
Posts: 320

For positive integer k, is the expression (k +
[#permalink]
Show Tags
21 Feb 2010, 14:40
For positive integer k, is the expression \((k + 2)(k^2 + 4k + 3)\) divisible by 4? (1) k is divisible by 8. (2) \(\frac{k + 1}{3}\) is an odd integer. SOURCE: Manhattan Tests Please explain. A little doubtful with this OA
_________________
Cheers! JT........... If u like my post..... payback in Kudos!!
Do not post questions with OAPlease underline your SC questions while postingTry posting the explanation along with your answer choice For CR refer Powerscore CR BibleFor SC refer Manhattan SC Guide
~~Better Burn Out... Than Fade Away~~



CEO
Joined: 17 Nov 2007
Posts: 3484
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth)  Class of 2011

Re: For positive integer k, is the expressio
[#permalink]
Show Tags
21 Feb 2010, 14:55
1) if k is divisible by 8 (and by 4 too), then k+2 is even but isn't divisible by 4 or 8. (k^2+4k+3)  an odd integer. > (even but not divisible by 4 or 8)*odd > the expression isn't divisible by 4. 2)k+1/3 is odd > k+1 is odd > k is even > k+2 is even. if k+2:  is divisible by 4, the expression is also divisible by 4  isn't divisible by 4, the expression isn't divisible by 4 insufficient
_________________
HOT! GMAT TOOLKIT 2 (iOS) / GMAT TOOLKIT (Android)  The OFFICIAL GMAT CLUB PREP APP, a musthave app especially if you aim at 700+  PrepGame



Senior Manager
Joined: 25 Jun 2009
Posts: 285

Re: For positive integer k, is the expressio
[#permalink]
Show Tags
22 Feb 2010, 12:57
jeeteshsingh wrote: For positive integer k, is the expression \((k + 2)(k^2 + 4k + 3)\) divisible by 4? (1) k is divisible by 8. (2) \(\frac{k + 1}{3}\) is an odd integer. SOURCE: Manhattan Tests Please explain. A little doubtful with this OA Another way of looking at it .. the expression \((k + 2)(k^2 + 4k + 3)\) can be written as (K+2)(K+1)(K+3) which means they are consecutive numbers. St1. K is divisible by 8 lets say K = 8 then the expression becomes 9*10*11 which is not divisible by 4, ( which holds for any value of K which is a multiple of 8) Hence Sufficient St 2 \(\frac{k + 1}{3}\) is an odd integer. Which K + 1 is an odd integer lets take K=2 If K =2 then the expression is divisible by 4 but not lets assume K = 8 then then again the expression is not divisible by 4 Hence Insufficient.



Intern
Joined: 03 Mar 2010
Posts: 37

Re: For positive integer k, is the expressio
[#permalink]
Show Tags
12 Mar 2010, 08:30
IMO A expression = (k+2)(k+1)(k+3) (1): (k+1)(k+3) is odd integer, (k+2) is only divisible by 2 but not by 4 so (1) is sufficient! (2): k=2 then expression=4*odd integer is divisible by 4 k=5 then expression=6*odd integer is not divisible by 4
_________________
Hardworkingly, you like my post, so kudos me.



Senior Manager
Status: Can't give up
Joined: 20 Dec 2009
Posts: 265

Re: For positive integer k, is the expressio
[#permalink]
Show Tags
12 Mar 2010, 11:59
clearly A.
B) when k=8 8+1/3 = divisible; when k=14 14+1/3 not divisible. Insuff



Intern
Joined: 19 Jun 2010
Posts: 23

Re: For positive integer k, is the expressio
[#permalink]
Show Tags
13 Jul 2010, 04:56
Another way to look at it: i)k is divisible by 8, so k is also divisible by 4. From the property of consecutive multiple , we know that the next term that is divisible by 4 after k is (k+4) (that is every 4th number from k). That means (K+2)(K+1)(K+3) is not divisible by 4.
ii) k+2 is even, but we don't know if it is a multiple of 4 or not. Plug in 1 and 3 for k+1 yields different answer



Manager
Joined: 04 Jan 2014
Posts: 112

Re: For positive integer k, is the expression (k + 2)(k2 + 4k +
[#permalink]
Show Tags
21 Apr 2014, 23:56
IMO D is the answer.
For second statement, if K + 1/3 or (K+1)/3, K is even. And so, in (K + 2) (K^2 + 4K + 3) = (K+2)(K+3)(K+1) = even x odd x odd
Am I correct?



SVP
Joined: 06 Sep 2013
Posts: 1852
Concentration: Finance

Re: For positive integer k, is the expression (k + 2)(k2 + 4k +
[#permalink]
Show Tags
29 Apr 2014, 16:52
Bunuel wrote: pretzel wrote: IMO D is the answer.
For second statement, if K + 1/3 or (K+1)/3, K is even. And so, in (K + 2) (K^2 + 4K + 3) = (K+2)(K+3)(K+1) = even x odd x odd
Am I correct? No, the correct answer is A. For positive integer k, is the expression (k + 2)(k^2 + 4k + 3) divisible by 4?\((k + 2)(k^2 + 4k + 3)=(k+1)(k+2)(k+3)\), so the expression is the product of three consecutive integers. (1) k is divisible by 8 > \(k=8n=even\) > \((k+1)(k+2)(k+3)=odd*even*odd\). Now, \(k+2=8n+2\), though even, is not a multiple of 8 (it's 2 greater than a multiple of 8), therefore the expression is not divisible by 8. Sufficient. (2) (k + 1)/3 is an odd integer > \(k+1=3*odd=odd\) > \(k=even\) > \((k+1)(k+2)(k+3)=odd*even*odd\). Now, \(k+2=even\) may or may not be divisible by 8, therefore the expression may or may not be divisible by 8. For example, consider \(k=2\) and \(k=6\). Not sufficient. Answer: A. Hope it's clear. I think that we are asked if the product of the three consecutive integers is divisible by 4 not by 8 Let me know Cheers! J



Math Expert
Joined: 02 Sep 2009
Posts: 47898

Re: For positive integer k, is the expression (k + 2)(k2 + 4k +
[#permalink]
Show Tags
30 Apr 2014, 07:38
jlgdr wrote: Bunuel wrote: pretzel wrote: IMO D is the answer.
For second statement, if K + 1/3 or (K+1)/3, K is even. And so, in (K + 2) (K^2 + 4K + 3) = (K+2)(K+3)(K+1) = even x odd x odd
Am I correct? No, the correct answer is A. For positive integer k, is the expression (k + 2)(k^2 + 4k + 3) divisible by 4?\((k + 2)(k^2 + 4k + 3)=(k+1)(k+2)(k+3)\), so the expression is the product of three consecutive integers. (1) k is divisible by 8 > \(k=8n=even\) > \((k+1)(k+2)(k+3)=odd*even*odd\). Now, \(k+2=8n+2\), though even, is not a multiple of 8 (it's 2 greater than a multiple of 8), therefore the expression is not divisible by 8. Sufficient. (2) (k + 1)/3 is an odd integer > \(k+1=3*odd=odd\) > \(k=even\) > \((k+1)(k+2)(k+3)=odd*even*odd\). Now, \(k+2=even\) may or may not be divisible by 8, therefore the expression may or may not be divisible by 8. For example, consider \(k=2\) and \(k=6\). Not sufficient. Answer: A. Hope it's clear. I think that we are asked if the product of the three consecutive integers is divisible by 4 not by 8 Let me know Cheers! J It should have been 4 instead of 8...
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Intern
Joined: 14 Sep 2009
Posts: 1

Re: For positive integer k, is the expression (k + 2)(k^2 + 4k +
[#permalink]
Show Tags
09 May 2014, 09:31
Hi Bunuel,
In statement 2, if we use k=6 then it does not meet the criteria (k+1)/3= Odd integer. However, if we use k=2 or k=8 or k=20 etc which meet the criteria, then we are able to say that statement 2 is sufficient. Is this line of thinking correct?



Intern
Joined: 17 Jul 2013
Posts: 38
Location: India
WE: Information Technology (Health Care)

Re: For positive integer k, is the expression (k + 2)(k^2 + 4k +
[#permalink]
Show Tags
29 Jul 2014, 11:20
(1): k is divisible by 8: so k is even in the expression (k + 2) (k^2 + 4k + 3), k^2 + 4k + 3 is odd if k =8 how? k^2 = 8^2 = even, 4k = even, 3 = odd so even + even + odd = odd and odd is never divisible by 4 it is sufficient if we test only k+2 is divisible by 4 since k is a multiple of 8, k is also a multiple of 4 (factor foundation rule) so a multiple of 4 + 2 is not divisible by 4. Sufficient (2) (k+1) / 3 = odd k + 1 = 3* odd = odd => k = odd  1 = even if k = 2 the expression is divisible by 4 if k = 4 the expression is not divisble by 4. Not sufficient Ans: A
_________________
I'm on 680... 20 days to reach 700 +



Manager
Joined: 22 Jan 2014
Posts: 172
WE: Project Management (Computer Hardware)

Re: For positive integer k, is the expression (k + 2)(k^2 + 4k +
[#permalink]
Show Tags
01 Feb 2015, 05:14
kishankolli 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 simplifying the expression > (k^3 + 6k^2 + 11k + 6) 1) k = 8p (say) => [(8p)^3 + 6*(8p)^2 + 88p + 6] mod 4 = 2 Not div. Hence sufficient. 2) (k+1)/3 = odd ==> k = even = 2p (say) => [8p^3 + 24p^2 + 22p + 6] mod 4 => 2(p+1) mod 4 if p is even then no ; if p is odd then yes Insufficient Hence, A.
_________________
Illegitimi non carborundum.



Intern
Joined: 08 Feb 2015
Posts: 1
Concentration: General Management, Entrepreneurship
GMAT Date: 08202015

Re: For positive integer k, is the expression (k + 2)(k^2 + 4k +
[#permalink]
Show Tags
25 Nov 2016, 09:34
thank for your explanation.



Intern
Joined: 18 May 2016
Posts: 27

Re: For positive integer k, is the expression (k + 2)(k^2 + 4k +
[#permalink]
Show Tags
20 Dec 2016, 08:58
Simplify to: (K+1)(K+2)(K+3). This expression will not be divisible by 4 if k is a multiple of 4: factors of 4 repeat once every 4 consecutive integers, thus in between no factors of 4. Stat1 gives you k=8q. So k=4*2q, k has 4 as a factor. Suff. Stat2 does not tell you anything about factors of 4 in the expression. Insuff. A



BSchool Forum Moderator
Joined: 28 Mar 2017
Posts: 1095
Location: India
Concentration: Finance, Technology
GPA: 4

Re: For positive integer k, is the expression (k + 2)(k^2 + 4k +
[#permalink]
Show Tags
22 Aug 2017, 08:49
kishankolli 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 Factorise the expression and we get > (k+1)(k+2)(k+3) If k is a multiple of 4 then (k+1)(k+2)(k+3) won't be divisible by 4, else for every nonmultiple of 4 value of k (k+1)(k+2)(k+3) will be divisible by 4. Statement A > k is divisible by 8; that means it is divisible by 4 also. Thus (k+1)(k+2)(k+3) will never be divisible by 4. SUFFICIENTStatement B > (k+1)/3 = 2n+1 => k+1=6n+3 =>k=6n+2 Now, for n=1; k=8 > (k+1)(k+2)(k+3) won't be divisible by 4. Now, for n=2; k=14 > (k+1)(k+2)(k+3) will be divisible by 4. Thus INSUFFICIENTAnswer: A
_________________
Kudos if my post helps!
Long And A Fruitful Journey  V21 to V41; If I can, So Can You!! Preparing for RC my way My study resources:1. Useful Formulae, Concepts and TricksQuant2. eGMAT's ALL SC Compilation3. LSAT RC compilation4. Actual LSAT CR collection by Broal5. QOTD RC (Carcass)6. Challange OG RC7. GMAT Prep Challenge RC




Re: For positive integer k, is the expression (k + 2)(k^2 + 4k + &nbs
[#permalink]
22 Aug 2017, 08:49



Go to page
1 2
Next
[ 26 posts ]



