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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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New post Updated on: 09 Jul 2013, 08:58
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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, 21:19.
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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New post 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.

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

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New post 09 Jul 2013, 20:33
1
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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New post 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.

Answer: E.


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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Re: PS - Integers and divisors  [#permalink]

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New post 28 Oct 2013, 00:24
2
3
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.

Answer: E.


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

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New post 20 Mar 2019, 02:09
Hi Bunuel,

As long as n^3 is a multiple of 8. n^3+8 / n , will always be an integer.

Tried picking numbers 2, 4, 8, 16, 32 & 64.

Hence there can more than 4 values of n.

Please correct me if i am wrong.
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Re: For how many different positive integers n is a divisor of n  [#permalink]

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New post 20 Mar 2019, 02:15
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Re: For how many different positive integers n is a divisor of n   [#permalink] 20 Mar 2019, 02:15
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