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Updated on: 09 Jul 2013, 08:58
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.
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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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
45%
(medium)
Question Stats:
61% (01:30) correct
39%
(01:39)
wrong
based on 793
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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
A. None
B. One
C. Two
D. Three
E. Four
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.
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.
28 Oct 2013, 00:24
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aakrity
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.
