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01 May 2024, 01:30
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E
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If \(n\) is an integer such that \(\frac{1}{6} < \frac{1}{n-1} < \frac{1}{3}\) and \((n - 5)(n - 3) \neq 0\), what is the value of \(n\)?
A. 2
B. 3
C. 4
D. 5
E. 6
A. 2
B. 3
C. 4
D. 5
E. 6
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05 May 2024, 16:45
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Official Solution:
If \(n\) is an integer such that \(\frac{1}{6} < \frac{1}{n-1} < \frac{1}{3}\) and \((n - 5)(n - 3) \neq 0\), what is the value of \(n\)?
A. 2
B. 3
C. 4
D. 5
E. 6
Given that \(n\) is an integer, for the inequality \(\frac{1}{6} < \frac{1}{n-1} < \frac{1}{3}\) to hold true, \(\frac{1}{n-1}\) must be either \(\frac{1}{5}\) or \(\frac{1}{4}\). Hence, \(n\) must be either 6 or 5.
From \((n - 5)(n - 3) \neq 0\), we can deduce that \(n\) cannot be \(5\) or \(3\). Therefore, since \(n\) is not 5, it must be 6.
Answer: E
If \(n\) is an integer such that \(\frac{1}{6} < \frac{1}{n-1} < \frac{1}{3}\) and \((n - 5)(n - 3) \neq 0\), what is the value of \(n\)?
A. 2
B. 3
C. 4
D. 5
E. 6
Given that \(n\) is an integer, for the inequality \(\frac{1}{6} < \frac{1}{n-1} < \frac{1}{3}\) to hold true, \(\frac{1}{n-1}\) must be either \(\frac{1}{5}\) or \(\frac{1}{4}\). Hence, \(n\) must be either 6 or 5.
From \((n - 5)(n - 3) \neq 0\), we can deduce that \(n\) cannot be \(5\) or \(3\). Therefore, since \(n\) is not 5, it must be 6.
Answer: E
General Discussion
01 May 2024, 01:57
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Bunuel
\(\frac{1}{6} < \frac{1}{n-1} < \frac{1}{3}\)
Taking the reciprocal of the equation
\(3 < n - 1 < 6\)
Adding 1 throughout the inequality
\(3+1 < n - 1 + 1 < 6 + 1\)
\(4 < n < 7\)
Hence, possible values of n = {5, 6}
\((n - 5)(n - 3) \neq 0\)
Inference: \(n\neq 5\) and \(n\neq 3\)
Therefore only possible value of n = 6
Option E
