Aug 22 09:00 PM PDT  10:00 PM PDT What you'll gain: Strategies and techniques for approaching featured GMAT topics, and much more. Thursday, August 22nd at 9 PM EDT Aug 24 07:00 AM PDT  09:00 AM PDT Learn reading strategies that can help even nonvoracious reader to master GMAT RC Aug 25 09:00 AM PDT  12:00 PM PDT Join a FREE 1day verbal workshop and learn how to ace the Verbal section with the best tips and strategies. Limited for the first 99 registrants. Register today! Aug 25 08:00 PM PDT  11:00 PM PDT Exclusive offer! Get 400+ Practice Questions, 25 Video lessons and 6+ Webinars for FREE.
Author 
Message 
TAGS:

Hide Tags

Intern
Joined: 01 Nov 2016
Posts: 6
Location: United Arab Emirates
WE: Information Technology (Telecommunications)

If n is a positive integer, is n divisible by 2?
[#permalink]
Show Tags
Updated on: 05 May 2018, 00:54
Question Stats:
61% (02:24) correct 39% (02:41) wrong based on 96 sessions
HideShow timer Statistics
If n is a positive integer, is n divisible by 2? (1) 7n  8 is divisible by 20 (2) 3n^2 + 2n + 5 is a prime number
Official Answer and Stats are available only to registered users. Register/ Login.
Originally posted by 53698 on 22 Jan 2017, 00:04.
Last edited by Bunuel on 05 May 2018, 00:54, edited 1 time in total.
Edited the question.



Math Expert
Joined: 02 Aug 2009
Posts: 7755

If n is a positive integer, is n divisible by 2?
[#permalink]
Show Tags
22 Jan 2017, 00:29
53698 wrote: If n is a positive integer, is n divisible by 2?
(1) 7n  8 is divisible by 20 (2) 3n^2 + 2n + 5 is a prime number 7n in 7n8 should be even, 7 is odd so n has to be even.. Statement I is sufficient Prime number above 2 is always odd, so 3n^2+2n+5 should be odd... 5 is odd, 2n is even so 3*n^2 should be even for entire term to be odd.. Thus n is even Suff D
_________________



Intern
Joined: 13 Dec 2016
Posts: 42

Re: If n is a positive integer, is n divisible by 2?
[#permalink]
Show Tags
22 Jan 2017, 02:40
chetan2u wrote: 53698 wrote: if n is a positive integer, is n divisibel by 2?
1) 7n8 is divisble by 20 2) 3nsqr+2n+5 is a prime number Pl post in correct forum Otherwise ans is D.. 7n in 7n8 should be even, 7 is odd so n has to be even.. Statement I is sufficient Prime number above 1 is always odd, so 3n^2+2n+5 should be odd... 5 is odd, 2n is even so 3*n^2 should be even for entire term to be odd.. Thus n is even Suff D HI Chetan  Wanted to reconfirm the rationale behind the highlighted text... 2 is a Prime No. which is even. But since 3n^2+2n+5 has the term 5, we are considering that the entire term if prime will be greater than 3 which are all odd.



Math Expert
Joined: 02 Aug 2009
Posts: 7755

Re: If n is a positive integer, is n divisible by 2?
[#permalink]
Show Tags
22 Jan 2017, 03:30
grichagupta wrote: chetan2u wrote: 53698 wrote: if n is a positive integer, is n divisibel by 2?
1) 7n8 is divisble by 20 2) 3nsqr+2n+5 is a prime number Pl post in correct forum Otherwise ans is D.. 7n in 7n8 should be even, 7 is odd so n has to be even.. Statement I is sufficient Prime number above 1 is always odd, so 3n^2+2n+5 should be odd... 5 is odd, 2n is even so 3*n^2 should be even for entire term to be odd.. Thus n is even Suff D HI Chetan  Wanted to reconfirm the rationale behind the highlighted text... 2 is a Prime No. which is even. But since 3n^2+2n+5 has the term 5, we are considering that the entire term if prime will be greater than 3 which are all odd. Hi You are correct in the rationale. Since n is a positive integer, it will be ATLEAST 1... And when n is ATLEAST 1, the term 3n^2+2n+5 will be greater than 2... Min value will be 3*1^2+2*1+5=10.... Thus value will be more than One..
_________________



Intern
Joined: 11 Mar 2015
Posts: 32

Re: If n is a positive integer, is n divisible by 2?
[#permalink]
Show Tags
18 Apr 2018, 02:23
Here, 1) 7n8 is divisible by 20. Hence, the number (7n8) must be multiple of 20 that is divisible by 2. So Statement (1) sufficient... 2) according to the question, 3n^2+2n+5= prime number(2,3,5,7,11..........). when n=2, => 3(2^2)+2*2+5 = 12+4+5=21 (prime) When n=3,5,7,11......... then 3n^2+2n+5= not prime so, n=2 that is divisible by 2 Hnce, statement (2) is sufficient.
Therefore, Correct answer is D.



Intern
Joined: 21 Aug 2017
Posts: 17
Location: India
Concentration: Entrepreneurship, Technology
WE: Operations (Other)

Re: If n is a positive integer, is n divisible by 2?
[#permalink]
Show Tags
18 Apr 2018, 04:29
mamun2013@gmail.com wrote: Here, 1) 7n8 is divisible by 20. Hence, the number (7n8) must be multiple of 20 that is divisible by 2. So Statement (1) sufficient... 2) according to the question, 3n^2+2n+5= prime number(2,3,5,7,11..........). when n=2, => 3(2^2)+2*2+5 = 12+4+5=21 (prime) CORRECTION:21 is not a prime number When n=3,5,7,11......... then 3n^2+2n+5= not prime so, n=2 that is divisible by 2 Hnce, statement (2) is sufficient.
Therefore, Correct answer is D. Thank you for your response. Though your final answer is correct, I disagree with the reasoning and request you to correct me where I go wrong. I believe the correct answer is C (which is why i posted here to clear this doubt). So the question asks if n being a positive integer is divisible by 2. Statement (1) states that 7n8 is divisible by 20. Thus, 7n8 = 20/40/60/80... the only number which satisfies the equation is n=4 where 7(4)8=20 which is divisible by 20, the numbers n=1/2/3/5/6/7/8/9/10/11/12/13/14/15/16/17/18/19/20 and so on do not give a solution to 7n8=divisible by 20. Therefore, unique solution is 4. 4 is divisible by 2. 4 is an even integer. Statement (2) states that \(3n^2\)+2n+5 is a prime number. Since n is a positive integer, it cannot be 0 which would make it 5 which is prime. Thus, taking the lowest positive integer, the equation will be \(3(1)^2\)+2(1)+5 = 10. Since 10 is the lowest number possible, any value of n >1 will always lead to a prime number being odd. To test it, \(3(2)^2\)+2(2)+5=21 which is not prime but odd. \(3(3)^3\)+2(3)=5=38. \(3(4)^2\)+2(4)+5=61 which is prime and odd. 4 is divisible by 2. 4 is even. 4 fits into the equation as well. However, Since only the number 4 is common in both statements (1) and (2), we can get the correct unique solution by combining both. Therefore, I think the answer is C. I want to understand why I am wrong. Thanks in advance for the help.



Retired Moderator
Joined: 22 Aug 2013
Posts: 1434
Location: India

Re: If n is a positive integer, is n divisible by 2?
[#permalink]
Show Tags
18 Apr 2018, 05:05
demblr wrote: mamun2013@gmail.com wrote: Here, 1) 7n8 is divisible by 20. Hence, the number (7n8) must be multiple of 20 that is divisible by 2. So Statement (1) sufficient... 2) according to the question, 3n^2+2n+5= prime number(2,3,5,7,11..........). when n=2, => 3(2^2)+2*2+5 = 12+4+5=21 (prime) CORRECTION:21 is not a prime number When n=3,5,7,11......... then 3n^2+2n+5= not prime so, n=2 that is divisible by 2 Hnce, statement (2) is sufficient.
Therefore, Correct answer is D. Thank you for your response. Though your final answer is correct, I disagree with the reasoning and request you to correct me where I go wrong. I believe the correct answer is C (which is why i posted here to clear this doubt). So the question asks if n being a positive integer is divisible by 2. Statement (1) states that 7n8 is divisible by 20. Thus, 7n8 = 20/40/60/80... the only number which satisfies the equation is n=4 where 7(4)8=20 which is divisible by 20, the numbers n=1/2/3/5/6/7/8/9/10/11/12/13/14/15/16/17/18/19/20 and so on do not give a solution to 7n8=divisible by 20. Therefore, unique solution is 4. 4 is divisible by 2. 4 is an even integer. Statement (2) states that \(3n^2\)+2n+5 is a prime number. Since n is a positive integer, it cannot be 0 which would make it 5 which is prime. Thus, taking the lowest positive integer, the equation will be \(3(1)^2\)+2(1)+5 = 10. Since 10 is the lowest number possible, any value of n >1 will always lead to a prime number being odd. To test it, \(3(2)^2\)+2(2)+5=21 which is not prime but odd. \(3(3)^3\)+2(3)=5=38. \(3(4)^2\)+2(4)+5=61 which is prime and odd. 4 is divisible by 2. 4 is even. 4 fits into the equation as well. However, Since only the number 4 is common in both statements (1) and (2), we can get the correct unique solution by combining both. Therefore, I think the answer is C. I want to understand why I am wrong. Thanks in advance for the help. Hello Statement 1: You have found a solution for n as n=4; that is not the only solution. Another solution would be n=24, another solution would be n=44.. but that is not important. What is important is that all values of n which satisfy this will be Even. Hence n is even. This statement is Sufficient. Statement 2: 3n^2 + 2n + 5 is prime. Now 5 is an odd prime, if we add an odd number to 5, it will always result in an even number greater than 5, which Cannot be prime. Thus 3n^2 + 2n must be even in order for 3n^2 + 2n + 5 to be a prime number. Now in 3n^2 + 2n, 2n is anyway even, hence we need 3n^2 to be also even so that the sum 3n^2 + 2n + 5 is odd. So no matter which value of n, that value will be even. Eg, n=4 makes it 3*4^2 + 2*4 + 5 = 61, which is prime. Now how many values of n satisfy this is not important, we know that any value of n which satisfies this will be even only. This statement is also Sufficient. When one or more statements is sufficient to answer a DS question, we do NOT combine the two statements. So the answer to this question is D, as each statement alone is sufficient to answer the question asked.



Intern
Joined: 21 Aug 2017
Posts: 17
Location: India
Concentration: Entrepreneurship, Technology
WE: Operations (Other)

Re: If n is a positive integer, is n divisible by 2?
[#permalink]
Show Tags
18 Apr 2018, 09:00
amanvermagmat wrote: demblr wrote: mamun2013@gmail.com wrote: Here, 1) 7n8 is divisible by 20. Hence, the number (7n8) must be multiple of 20 that is divisible by 2. So Statement (1) sufficient... 2) according to the question, 3n^2+2n+5= prime number(2,3,5,7,11..........). when n=2, => 3(2^2)+2*2+5 = 12+4+5=21 (prime) CORRECTION:21 is not a prime number When n=3,5,7,11......... then 3n^2+2n+5= not prime so, n=2 that is divisible by 2 Hnce, statement (2) is sufficient.
Therefore, Correct answer is D. Thank you for your response. Though your final answer is correct, I disagree with the reasoning and request you to correct me where I go wrong. I believe the correct answer is C (which is why i posted here to clear this doubt). So the question asks if n being a positive integer is divisible by 2. Statement (1) states that 7n8 is divisible by 20. Thus, 7n8 = 20/40/60/80... the only number which satisfies the equation is n=4 where 7(4)8=20 which is divisible by 20, the numbers n=1/2/3/5/6/7/8/9/10/11/12/13/14/15/16/17/18/19/20 and so on do not give a solution to 7n8=divisible by 20. Therefore, unique solution is 4. 4 is divisible by 2. 4 is an even integer. Statement (2) states that \(3n^2\)+2n+5 is a prime number. Since n is a positive integer, it cannot be 0 which would make it 5 which is prime. Thus, taking the lowest positive integer, the equation will be \(3(1)^2\)+2(1)+5 = 10. Since 10 is the lowest number possible, any value of n >1 will always lead to a prime number being odd. To test it, \(3(2)^2\)+2(2)+5=21 which is not prime but odd. \(3(3)^3\)+2(3)=5=38. \(3(4)^2\)+2(4)+5=61 which is prime and odd. 4 is divisible by 2. 4 is even. 4 fits into the equation as well. However, Since only the number 4 is common in both statements (1) and (2), we can get the correct unique solution by combining both. Therefore, I think the answer is C. I want to understand why I am wrong. Thanks in advance for the help. Hello Statement 1: You have found a solution for n as n=4; that is not the only solution. Another solution would be n=24, another solution would be n=44.. but that is not important. What is important is that all values of n which satisfy this will be Even. Hence n is even. This statement is Sufficient. Statement 2: 3n^2 + 2n + 5 is prime. Now 5 is an odd prime, if we add an odd number to 5, it will always result in an even number greater than 5, which Cannot be prime. Thus 3n^2 + 2n must be even in order for 3n^2 + 2n + 5 to be a prime number. Now in 3n^2 + 2n, 2n is anyway even, hence we need 3n^2 to be also even so that the sum 3n^2 + 2n + 5 is odd. So no matter which value of n, that value will be even. Eg, n=4 makes it 3*4^2 + 2*4 + 5 = 61, which is prime. Now how many values of n satisfy this is not important, we know that any value of n which satisfies this will be even only. This statement is also Sufficient. When one or more statements is sufficient to answer a DS question, we do NOT combine the two statements. So the answer to this question is D, as each statement alone is sufficient to answer the question asked. Hi, thank you for the detailed explanation. I need a clarification with the following: In statement 2: how did you get to that thought process? I first thought of using numbers as examples and working it out. Secondly,in statement 2 itself, how does one know that no matter which value of n we consider, it will be even. I was thinking if we take n as 3, then 3^2 will be 9, 9*3 will be 27 which makes 3n^2 odd. Can you please help me understand how you arrived at the conclusion that statement 2 is sufficient as well? Thanks in advance for the help.



Retired Moderator
Joined: 22 Aug 2013
Posts: 1434
Location: India

Re: If n is a positive integer, is n divisible by 2?
[#permalink]
Show Tags
18 Apr 2018, 10:58
[/quote]
Hi, thank you for the detailed explanation. I need a clarification with the following: In statement 2: how did you get to that thought process? I first thought of using numbers as examples and working it out. Secondly,in statement 2 itself, how does one know that no matter which value of n we consider, it will be even. I was thinking if we take n as 3, then 3^2 will be 9, 9*3 will be 27 which makes 3n^2 odd. Can you please help me understand how you arrived at the conclusion that statement 2 is sufficient as well?
Thanks in advance for the help.[/quote]
Hello
I have used the basic properties of numbers here. 3n^2 + 2n + 5 is what is given to us. Now 5 is odd, and adding 3n^2 + 2n to 5 should give us a prime number, which obviously will be greater than 5. Now all prime numbers above 5 are odd only (except 2 all prime numbers are odd). So this means 3n^2 + 2n + 5 should be odd.
Now 5 is odd, and adding odd to odd gives us even number. So 3n^2 + 2n must be even, and not odd. Now 2n will always be even, because 2 multiplied by any integer will give an even result only. So we have 3n^2 + 2n as even, where 2n is already even.
Now if we add even with odd, we will get an odd result (which we dont want here), but if we add even with even, we will get even result. Since 2n is already even, 3n^2 must also be even (for their sum to be even)... And 3 is an odd number, odd multiplied by odd gives odd, while odd multiplied by even gives even. So for 3n^2 to be even, n^2 also must be even. And n^2 will be even only when n is even.
Thats how I concluded that n has to be even.



Manager
Joined: 28 Nov 2017
Posts: 142
Location: Uzbekistan

Re: If n is a positive integer, is n divisible by 2?
[#permalink]
Show Tags
26 Apr 2018, 04:50
Tridhipal wrote: If n is a positive integer, is n divisible by 2?
1) 7n8 is divisible by 20 2) \(3n^{2}+2n+5\) is a prime number. (1) To be divisible by \(20\), \(7n8\) must be even number. Hence \(n\) is even and divisible by \(2\). Suff. (2) If \(n\) is odd, then the value of expression is even number (larger than \(2\)), hence is not prime. Therefore, \(n\) must be even number. Suff. Answer: D
_________________



Senior Manager
Joined: 21 Jan 2015
Posts: 458
Location: India
Concentration: Strategy, Marketing
GMAT 1: 620 Q48 V28 GMAT 2: 690 Q49 V35
WE: Sales (Consumer Products)

Re: If n is a positive integer, is n divisible by 2?
[#permalink]
Show Tags
26 Apr 2018, 07:09
Tridhipal wrote: If n is a positive integer, is n divisible by 2?
1) 7n8 is divisible by 20 2) \(3n^{2}+2n+5\) is a prime number. Ans: D Stat 1) 7n8 is divisible by 20 means, 7n8 = 20k ; n = (20k+8)/7 here we can take 2 common and as the denominator is 7 we know that 2 will remain so n is even(Sufficient) Stat 2) for solving this equation we will take the n as positive int and put the value in the equation. We will get that when ever n is even we will get the Prime number.. (Suff) Ans: D
_________________
 The Mind is Everything, What we Think we Become.



Director
Joined: 02 Oct 2017
Posts: 726

Re: If n is a positive integer, is n divisible by 2?
[#permalink]
Show Tags
03 May 2018, 12:45
I) every multiple of 20 is even So for 7n8 to divide by 20,7n8 need to be even.So n has to be odd. So n is not divisible by 2. Sufficient II)3n^2 +2n+5 is prime All prime except 2 are odd.above number cannot be 2 as n is positive integer so n cannot be zero. For above to be odd N has to be odd.so n is not divisible by 2 Sufficient D is answer Posted from my mobile device
_________________
Give kudos if you like the post



Founder
Joined: 04 Dec 2002
Posts: 17979
Location: United States (WA)
GPA: 3.5

Re: If n is a positive integer, is n divisible by 2?
[#permalink]
Show Tags
04 May 2018, 21:04
Moved topic to the correct forum from the GENERAL GMAT forum. Unsure about the source of this question. Thx.
_________________
Founder of GMAT Club Just starting out with GMAT? Start here... Want to know application stats & Profiles from last year? Check the Decision Tracker



Target Test Prep Representative
Status: Head GMAT Instructor
Affiliations: Target Test Prep
Joined: 04 Mar 2011
Posts: 2819

Re: If n is a positive integer, is n divisible by 2?
[#permalink]
Show Tags
01 Jun 2018, 11:32
53698 wrote: If n is a positive integer, is n divisible by 2?
(1) 7n  8 is divisible by 20 (2) 3n^2 + 2n + 5 is a prime number In order for an integer n to be divisible by 2, n must be even. So we need to determine whether n is even. Statement One Alone: 7n8 is divisible by 20. That is, 7n  8 = 20k for some integer k. So we have: 7n = 20k + 8 n = 4(5k + 2)/7 Since n is an integer and 4 and 7 have no common factor other than 1, then 5k + 2 must be divisible by 7. In other words, (5k + 2)/7 is an integer. Thus, we see that 4(5k + 2)/7 is the same as 4 x integer, which will always be even, so n will always be even. Statement one alone is sufficient to answer the question. Statement Two Alone: 3n^2+2n+5 is a prime number. Notice that since n is a positive integer, 3n^2+2n+5 must be greater than 2. If 3n^2 + 2n + 5 is a prime, then it must be an odd prime and the only way 3n^2+2n+5 is an odd prime is if n is even. Statement two alone is also sufficient to answer the question. Answer: D
_________________
5star rated online GMAT quant self study course See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews
If you find one of my posts helpful, please take a moment to click on the "Kudos" button.



NonHuman User
Joined: 09 Sep 2013
Posts: 12052

Re: If n is a positive integer, is n divisible by 2?
[#permalink]
Show Tags
11 Aug 2019, 09:48
Hello from the GMAT Club BumpBot! Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up  doing my job. I think you may find it valuable (esp those replies with Kudos). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________




Re: If n is a positive integer, is n divisible by 2?
[#permalink]
11 Aug 2019, 09:48






