M04-15

Official Solution:


If \(p\) and \(q\) are nonzero integers and \(n=\frac{p}{q}\), is \(n\) an integer?

(1) \(n^2\) is an integer.

For \(n^2\) to be an integer, \(n\) must either be an integer or an irrational number such as \(\sqrt{3}\) (it's important to note that \(n\) cannot be a reduced fraction like \(\frac{2}{3}\) or \(\frac{11}{3}\) because in these cases \(n^2\) would not yield an integer). However, since \(n\) can be represented as the ratio of two integers (i.e., \(n=\frac{p}{q}\)), it cannot be an irrational number (by definition, an irrational number is any real number that cannot be expressed as a fraction \(\frac{a}{b}\), where \(a\) and \(b\) are integers). Consequently, the only remaining possibility is that \(n\) is an integer. This is sufficient.

(2) \(\frac{2n+4}{2}\) is an integer.

The above gives \(\frac{2n+4}{2}=n+2 =\text{integer}\). From this it follows that \(n\) must be an integer. Sufficient.


Answer: D
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Good question. I was wondering why do we have n=P/Q with both p,q = non zero integers, but this information is useful to evaluate statement 1.

Per statement 1, \(n^2\) = integer. Now \(n^2=4\) or \(n^2 =2\) but \(n^2\)=not possible as this will give n = \(\sqrt{2}\) but as n=P/Q (ratio of integers), it can not be equal to \(\sqrt{2}\). For n = \(\sqrt{2}\) = \(\sqrt{2}/1\) ----> this means that P = \(\sqrt{2}\) and this will violate the information in the question that P is an integer. Thus this statement is sufficient.

Per statement 2, 2n+4 / 2 = integer ---> n+2 = integer ---> n = integer - 2 = another integer. Thus this statement is sufficient as well.

Thus D is the correct answer.
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What if n^2=15? n would equal root(15), otherwise stated as root(30/2) which is not an integer but satisfies the conditions. Please help!

n is a rational number, it's a ratio of two integers and thus it cannot equal to irrational number such as \(\sqrt{15}\). \(\sqrt{15}\) cannot be expressed as the ratio of two integers. BTW this is explained in the solution.
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I think this is a high-quality question and I agree with explanation. Great Question !!

I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
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I think this is a high-quality question and I agree with explanation.
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