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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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