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

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

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27 Oct 2010, 21:19
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For how many different positive integers n is a divisor of n^3 + 8?

A. None
B. One
C. Two
D. Three
E. Four
[Reveal] Spoiler: OA

Last edited by Bunuel on 09 Jul 2013, 08:58, edited 1 time in total.
Renamed the topic and edited the question.
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Re: PS - Integers and divisors [#permalink]

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27 Oct 2010, 21:25
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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.

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Kudos [?]: 34 [1] , given: 84

Re: For how many different positive integers n is a divisor of n [#permalink]

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09 Jul 2013, 20:33
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n^3 + 8 = (n+2)(n^2-2n+4)
--> 2 divisors possible, plus 1 and n
we have 4 divisors.
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Re: PS - Integers and divisors [#permalink]

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27 Oct 2013, 09:56
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 ?
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Posts: 34438
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Kudos [?]: 79517 [2] , given: 10018

Re: PS - Integers and divisors [#permalink]

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28 Oct 2013, 00:24
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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.
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Re: For how many different positive integers n is a divisor of n [#permalink]

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23 Nov 2014, 12:13
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Re: For how many different positive integers n is a divisor of n [#permalink]

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13 Dec 2015, 14:52
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).

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Re: For how many different positive integers n is a divisor of n   [#permalink] 13 Dec 2015, 14:52
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