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05 Sep 2022, 05:21
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C
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Difficulty:
5%
(low)
Question Stats:
94% (00:51) correct
6%
(01:20)
wrong
based on 63
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For which of the following values of n is \(\frac{96+n}{n}\) NOT an integer?
(A) 4
(B) 8
(C) 9
(D) 12
(E) 16
(A) 4
(B) 8
(C) 9
(D) 12
(E) 16
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05 Sep 2022, 08:21
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Prime factorising 96: \(2^5 * 3\)
Rewriting the question using the prime factorisation from above: \(\frac{2^5*3 + n}{n}\)
Given the addition sign between the numerators, the only way to divide is to take out a common factor. Therefore, \(n\) needs to be a number which can be made using any combination of the five \(2\)'s and single \(3\) in order for there to be a common factor.
The only answer choice which has a number which cannot be made using \(2^5\) and \(3\), is answer choice C.
\(\frac{2^5*3 + (9)}{(9)}\)
\(\frac{2^5*3 + 3^2 }{ 3^2}\)
\(\frac{3(2^5+3)}{3^2}\)
\(\frac{(2^5+3)}{3}\)
Answer C
Rewriting the question using the prime factorisation from above: \(\frac{2^5*3 + n}{n}\)
Given the addition sign between the numerators, the only way to divide is to take out a common factor. Therefore, \(n\) needs to be a number which can be made using any combination of the five \(2\)'s and single \(3\) in order for there to be a common factor.
The only answer choice which has a number which cannot be made using \(2^5\) and \(3\), is answer choice C.
\(\frac{2^5*3 + (9)}{(9)}\)
\(\frac{2^5*3 + 3^2 }{ 3^2}\)
\(\frac{3(2^5+3)}{3^2}\)
\(\frac{(2^5+3)}{3}\)
Answer C
05 Sep 2022, 09:04
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Key (and often-tested) property: \(\frac{a+b}{c} = \frac{a}{c} + \frac{b}{c}\)
So, \(\frac{96+n}{n} = \frac{96}{n} + \frac{n}{n}= \frac{96}{n} + 1\)
Since 1 is an integer, the question boils down to finding a value of n so that \(\frac{96}{n}\) is NOT an integer
Since 96 is not the visible by 9, we know that \(\frac{96}{n}\) will not be an integer when n = 9
Answer: C
