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

Jay@ManhattanReview: GMAT Instructor; Posts: 3008; Joined: Mon Aug 22, 2016 6:19 am; Location: Grand Central / New York; Thanked: 470 times; Followed by:34 members

by Jay@ManhattanReview » Thu May 31, 2018 11:50 pm

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NandishSS wrote:If n is a positive integer, is n divisible by 2?

A)7n-8 is divisible by 20.
B)3n^2+2n+5 is a prime number.

OA: D

We are given that n is a positive integer.

We have to find out whether n is divisible by 2.

Let's take each statement one by one.

1) 7n - 8 is divisible by 20.

Since 7n - 8 is divisible by 20 (an even number), 7n - 8 must be even.

=> 7n - 8 = Even

7n = Even + 8 = Even

7*n = even

Since 7 is odd, n must be even. Thus, n divisible by 2. Sufficient.

2) 3n^2 + 2n + 5 is a prime number.

3n^2 + 2n + 5 = Odd; it cannot be 2 since 3n^2 + 2n + 5 must be greater than 5 and all the prime numbers greater than 5 are odd.

=> 3n^2 + 2n = Odd - 5

3n^2 + 2n = Odd - Odd

3n^2 + 2n = Even

3n^2 = Even - 2n

3n^2 = Even - Even

3n^2 = Even

Since 3 is odd, n^2 must be even, thereby n must be even. Thus, n divisible by 2. Sufficient.

The correct answer: D

Hope this helps!

-Jay
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Vincen: Legendary Member; Posts: 2898; Joined: Thu Sep 07, 2017 2:49 pm; Thanked: 6 times; Followed by:5 members

by Vincen » Thu May 31, 2018 11:59 pm

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Hello NandishSS.

I will try to solve your question.

We know that n is a positive integer and we want to know if n is divisible by 2.

First Statement

A)7n-8 is divisible by 20.

This statement tells us that $$7n-8=20\cdot k\ \ \ ,\ \ \ k\in\mathbb{Z}$$ $$\Rightarrow\ \ 7n=20\cdot k\ +8\ \ ,\ \ \ k\in\mathbb{Z}$$ $$\Rightarrow\ \ 7n=2\left(10\cdot k\ +4\right)\ \ ,\ \ \ k\in\mathbb{Z}$$ The last equation says that 7b is divisible by 2 and since 7 is not divisible by 2, we conclude that n is divisible by 2.

Hence, this statement is SUFFICIENT.

Second Statement

B)3n^2+2n+5 is a prime number.

Here we have to see the following:

- Since n is a positive integer then n>0 (n >= 1)
- If n>0 then 3n^2+2n+5>= 10.
- Since 3n^2+2n+5 is a prime number and it is greater or equal than 10, then 3n^2+2n+5 must be an odd number.

So, we have the following: $$3n^2+2n+5=odd\ \ \ \Rightarrow\ \ 3n^2+2n=odd-5\ \ \ \Rightarrow\ \ \ 3n^2+2n=even.$$ $$\Rightarrow\ \ 3n^2=even\ -\ 2n\ \ \ \ \Rightarrow\ \ \ 3n^2=even\ -even\ \ \ \Rightarrow\ \ \ 3n^2=even.$$ Since 3 is odd, then we have to n^2 must be even. This only happens when n is even.

Therefore, n is even and this implies that n is divisible by 2.

So, this statement is also SUFFICIENT.

In conclusion, each statement is SUFFICIENT, so the correct answer is the option D.

I hope it helps.

Regards.

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Jeff@TargetTestPrep: GMAT Instructor; Posts: 1462; Joined: Thu Apr 09, 2015 9:34 am; Location: New York, NY; Thanked: 39 times; Followed by:22 members

by Jeff@TargetTestPrep » Fri Jun 01, 2018 12:16 pm

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NandishSS wrote:If n is a positive integer, is n divisible by 2?

A)7n-8 is divisible by 20.
B)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.

Answer: D

Jeffrey Miller
Head of GMAT Instruction
[email protected]

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