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16 Sep 2014, 00:18
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If \(q\) is a positive integer, is \(p*\frac{q}{\sqrt{q} }\) an integer?
(1) \(q = p^2\)
(2) \(p\) is a positive integer
(1) \(q = p^2\)
(2) \(p\) is a positive integer
16 Sep 2014, 00:18
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Official Solution:
If \(q\) is a positive integer, is \(p*\frac{q}{\sqrt{q} }\) an integer?
(1) \(q = p^2\).
By taking the square root of both sides of the equation, we obtain \(\sqrt{q} = |p|\). Thus, \(\sqrt{q} = p\), if \(p\) is positive or \(\sqrt{q} = -p\), if \(p\) is negative. Substituting \(p\) with \(\sqrt{q}\) in the expression above results in \(\sqrt{q} * \frac{q}{\sqrt{q} } = q\) and substituting \(p\) with \(-\sqrt{q}\) in the expression above results in \(-\sqrt{q} * \frac{q}{\sqrt{q} } = -q\). As the expression simplifies to \(q\) or to \(-q\), and it is given that \(q\) is an integer, the expression is also an integer. Therefore, statement (1) is sufficient to answer the question.
(2) \(p\) is a positive integer.
We can rewrite the given expression as \(p*\frac{q}{\sqrt{q} }=p*\frac{(\sqrt{q})^2}{\sqrt{q} }=p*\sqrt{q}\). However, simply knowing that \(p\) is a positive integer is not enough to determine whether this expression is an integer or not. If \(q\) is a square of an integer, then \(\sqrt{q}\) is also an integer, and \(p*\sqrt{q}\) is an integer. However, if \(q\) is not a square of an integer, then \(\sqrt{q}\) is not an integer, and \(p*\sqrt{q}\) is not an integer. Thus, statement (2) alone is not sufficient to determine whether the expression is an integer or not.
Answer: A
If \(q\) is a positive integer, is \(p*\frac{q}{\sqrt{q} }\) an integer?
(1) \(q = p^2\).
By taking the square root of both sides of the equation, we obtain \(\sqrt{q} = |p|\). Thus, \(\sqrt{q} = p\), if \(p\) is positive or \(\sqrt{q} = -p\), if \(p\) is negative. Substituting \(p\) with \(\sqrt{q}\) in the expression above results in \(\sqrt{q} * \frac{q}{\sqrt{q} } = q\) and substituting \(p\) with \(-\sqrt{q}\) in the expression above results in \(-\sqrt{q} * \frac{q}{\sqrt{q} } = -q\). As the expression simplifies to \(q\) or to \(-q\), and it is given that \(q\) is an integer, the expression is also an integer. Therefore, statement (1) is sufficient to answer the question.
(2) \(p\) is a positive integer.
We can rewrite the given expression as \(p*\frac{q}{\sqrt{q} }=p*\frac{(\sqrt{q})^2}{\sqrt{q} }=p*\sqrt{q}\). However, simply knowing that \(p\) is a positive integer is not enough to determine whether this expression is an integer or not. If \(q\) is a square of an integer, then \(\sqrt{q}\) is also an integer, and \(p*\sqrt{q}\) is an integer. However, if \(q\) is not a square of an integer, then \(\sqrt{q}\) is not an integer, and \(p*\sqrt{q}\) is not an integer. Thus, statement (2) alone is not sufficient to determine whether the expression is an integer or not.
Answer: A
15 Dec 2014, 08:14
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Arnav180
The result would still be an integer: \(p=-\sqrt{q}\). Substitute \(p\): \(p\frac{q}{\sqrt{q}}=-\sqrt{q}*\frac{q}{\sqrt{q}} = -q = \text{integer}\). Sufficient.
