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Re: PS - Integers and divisors [#permalink]
27 Oct 2013, 23:24

2

This post received KUDOS

Expert's post

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.

Re: For how many different positive integers n is a divisor of n [#permalink]
23 Nov 2014, 11:13

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