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

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

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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

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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

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.
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Joined: 02 Sep 2009
Posts: 53020
Re: PS - Integers and divisors  [#permalink]

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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.

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

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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]

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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]

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

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20 Jan 2019, 22:15
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Re: For how many different positive integers n is a divisor of n   [#permalink] 20 Jan 2019, 22:15
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# For how many different positive integers n is a divisor of n

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