GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 20 Feb 2019, 14:29

### 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 February
PrevNext
SuMoTuWeThFrSa
272829303112
3456789
10111213141516
17181920212223
242526272812
Open Detailed Calendar
• ### Free GMAT Prep Hour

February 20, 2019

February 20, 2019

08:00 PM EST

09:00 PM EST

Strategies and techniques for approaching featured GMAT topics. Wednesday, February 20th at 8 PM EST

February 21, 2019

February 21, 2019

10:00 PM PST

11:00 PM PST

Kick off your 2019 GMAT prep with a free 7-day boot camp that includes free online lessons, webinars, and a full GMAT course access. Limited for the first 99 registrants! Feb. 21st until the 27th.

# For how many different positive integers n is a divisor of n

Author Message
TAGS:

### Hide Tags

Intern
Joined: 07 Feb 2009
Posts: 13
For how many different positive integers n is a divisor of n  [#permalink]

### Show Tags

Updated on: 09 Jul 2013, 07:58
1
10
00:00

Difficulty:

55% (hard)

Question Stats:

58% (01:34) correct 42% (01:36) wrong based on 457 sessions

### HideShow timer Statistics

For how many different positive integers n is a divisor of n^3 + 8?

A. None
B. One
C. Two
D. Three
E. Four

Originally posted by vivaslluis on 27 Oct 2010, 20:19.
Last edited by Bunuel on 09 Jul 2013, 07:58, edited 1 time in total.
Renamed the topic and edited the question.
Math Expert
Joined: 02 Sep 2009
Posts: 53020
Re: PS - Integers and divisors  [#permalink]

### Show Tags

27 Oct 2010, 20:25
9
2
For how many different positive integers n is a divisor of n^3 + 8?

A. None
B. One
C. Two
D. Three
E. Four

The question asks for how many different positive integers $$n$$, $$n^3+8$$ is divisible by $$n$$.

Well the first term $$n^3$$ is divisible by $$n$$, the second term, 8, to be divisible by $$n$$, should be a factor of 8.

$$8=2^3$$ hence it has 3+1=4 factors: 1, 2, 4, and 8. Therefore $$n^3+8$$ is divisible by $$n$$ for 4 values of $$n$$: 1, 2, 4, and 8.

_________________
##### General Discussion
Intern
Joined: 28 May 2012
Posts: 25
Concentration: Finance, General Management
GMAT 1: 700 Q50 V35
GPA: 3.28
WE: Analyst (Investment Banking)
Re: For how many different positive integers n is a divisor of n  [#permalink]

### Show Tags

09 Jul 2013, 19:33
1
n^3 + 8 = (n+2)(n^2-2n+4)
--> 2 divisors possible, plus 1 and n
we have 4 divisors.
Intern
Joined: 08 Oct 2011
Posts: 36
Re: PS - Integers and divisors  [#permalink]

### Show Tags

27 Oct 2013, 08:56
Bunuel wrote:
For how many different positive integers n is a divisor of n^3 + 8?

A. None
B. One
C. Two
D. Three
E. Four

The question asks for how many different positive integers $$n$$, $$n^3+8$$ is divisible by $$n$$.

Well the first term $$n^3$$ is divisible by $$n$$, the second term, 8, to be divisible by $$n$$, should be a factor of 8.

$$8=2^3$$ hence it has 3+1=4 factors: 1, 2, 4, and 8. Therefore $$n^3+8$$ is divisible by $$n$$ for 4 values of $$n$$: 1, 2, 4, and 8.

Bunnel at first I got the answer 4.

Then I plugged in n=3 and got zero. PS questions are supposed to have one right answer. Why are we using only $$2$$ here ?
Math Expert
Joined: 02 Sep 2009
Posts: 53020
Re: PS - Integers and divisors  [#permalink]

### Show Tags

27 Oct 2013, 23:24
2
2
aakrity wrote:
Bunuel wrote:
For how many different positive integers n is a divisor of n^3 + 8?

A. None
B. One
C. Two
D. Three
E. Four

The question asks for how many different positive integers $$n$$, $$n^3+8$$ is divisible by $$n$$.

Well the first term $$n^3$$ is divisible by $$n$$, the second term, 8, to be divisible by $$n$$, should be a factor of 8.

$$8=2^3$$ hence it has 3+1=4 factors: 1, 2, 4, and 8. Therefore $$n^3+8$$ is divisible by $$n$$ for 4 values of $$n$$: 1, 2, 4, and 8.

Bunnel at first I got the answer 4.

Then I plugged in n=3 and got zero. PS questions are supposed to have one right answer. Why are we using only $$2$$ here ?

You cannot pick arbitrary numbers here.

The question asks "for how many different positive integers n is a divisor of n^3 + 8", so for how many different n's $$\frac{n^3 + 8}{n}=integer$$.

Now, $$\frac{n^3 + 8}{n}=n^2+\frac{8}{n}$$ to be an integer n must be a factor of 8. 8 has 4 factors, thus for 4 values of n $$\frac{n^3 + 8}{n}$$ will be an integer: if n is 1, 2, 4, or 8.

Hope it's clear.
_________________
Non-Human User
Joined: 09 Sep 2013
Posts: 9869
Re: For how many different positive integers n is a divisor of n  [#permalink]

### Show Tags

20 Jan 2019, 22:15
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: For how many different positive integers n is a divisor of n   [#permalink] 20 Jan 2019, 22:15
Display posts from previous: Sort by