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# If n is a positive integer, is n divisible by 2?

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If n is a positive integer, is n divisible by 2?  [#permalink]

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Updated on: 05 May 2018, 00:54
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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

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.
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If n is a positive integer, is n divisible by 2?  [#permalink]

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22 Jan 2017, 00:29
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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 7n-8 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
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Re: If n is a positive integer, is n divisible by 2?  [#permalink]

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22 Jan 2017, 02:40
chetan2u wrote:
53698 wrote:
if n is a positive integer, is n divisibel by 2?

1) 7n-8 is divisble by 20
2) 3nsqr+2n+5 is a prime number

Pl post in correct forum
Otherwise ans is D..

7n in 7n-8 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.
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Re: If n is a positive integer, is n divisible by 2?  [#permalink]

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22 Jan 2017, 03:30
grichagupta wrote:
chetan2u wrote:
53698 wrote:
if n is a positive integer, is n divisibel by 2?

1) 7n-8 is divisble by 20
2) 3nsqr+2n+5 is a prime number

Pl post in correct forum
Otherwise ans is D..

7n in 7n-8 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..
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Re: If n is a positive integer, is n divisible by 2?  [#permalink]

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18 Apr 2018, 02:23
Here, 1) 7n-8 is divisible by 20. Hence, the number (7n-8) 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.

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Re: If n is a positive integer, is n divisible by 2?  [#permalink]

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18 Apr 2018, 04:29
mamun2013@gmail.com wrote:
Here, 1) 7n-8 is divisible by 20. Hence, the number (7n-8) 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.

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 7n-8 is divisible by 20. Thus, 7n-8 = 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 7n-8=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.
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Re: If n is a positive integer, is n divisible by 2?  [#permalink]

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18 Apr 2018, 05:05
demblr wrote:
mamun2013@gmail.com wrote:
Here, 1) 7n-8 is divisible by 20. Hence, the number (7n-8) 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.

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 7n-8 is divisible by 20. Thus, 7n-8 = 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 7n-8=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.
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Re: If n is a positive integer, is n divisible by 2?  [#permalink]

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18 Apr 2018, 09:00
amanvermagmat wrote:
demblr wrote:
mamun2013@gmail.com wrote:
Here, 1) 7n-8 is divisible by 20. Hence, the number (7n-8) 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.

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 7n-8 is divisible by 20. Thus, 7n-8 = 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 7n-8=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.
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Re: If n is a positive integer, is n divisible by 2?  [#permalink]

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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.
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Re: If n is a positive integer, is n divisible by 2?  [#permalink]

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26 Apr 2018, 04:50
Tridhipal 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.

(1) To be divisible by $$20$$, $$7n-8$$ 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.

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Re: If n is a positive integer, is n divisible by 2?  [#permalink]

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26 Apr 2018, 07:09
Tridhipal 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.

Ans: D

Stat 1) 7n-8 is divisible by 20 means, 7n-8 = 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
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Re: If n is a positive integer, is n divisible by 2?  [#permalink]

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03 May 2018, 12:45
I) every multiple of 20 is even

So for 7n-8 to divide by 20,7n-8 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

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Re: If n is a positive integer, is n divisible by 2?  [#permalink]

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04 May 2018, 21:04
1
Moved topic to the correct forum from the GENERAL GMAT forum. Unsure about the source of this question. Thx.
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Re: If n is a positive integer, is n divisible by 2?  [#permalink]

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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:

7n-8 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.

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