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The variable $q$ is an integer and a perfect square.
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20 Nov 2025, 09:18
20 Nov 2025, 09:18
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75% (01:32) correct
25% (00:53) wrong based on 4 sessions
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The variable $q$ is an integer and a perfect square. Which of the following is also an integer?
Indicate all such integers.
(A) $\(q^2+\sqrt{q}+1\)$
(B) $\(q^3+\frac{q}{2}+3\)$
(c) $\(\frac{\sqrt{q}}{2}\)$
(D) $\(q^2+q+\sqrt{q}\)$
Indicate all such integers.
(A) $\(q^2+\sqrt{q}+1\)$
(B) $\(q^3+\frac{q}{2}+3\)$
(c) $\(\frac{\sqrt{q}}{2}\)$
(D) $\(q^2+q+\sqrt{q}\)$
- Part of the project: GRE Daily Quant Challenge - Topic-Wise 2025!!!
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Re: The variable $q$ is an integer and a perfect square.
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11 Dec 2025, 02:28
11 Dec 2025, 02:28
Expert Reply
Let $\(q=k^2\)$ for some integer $k$ since $q$ is a perfect square.
- (A) $\(q^2+\sqrt{q}+1=k^4+k+1\)$, which is an integer for any integer $k$.
- (B) $\(q^3+\frac{q}{2}+3=k^6+\frac{k^2}{2}+3\)$. This is an integer only if $\(k^2\)$ is even, i.e., $k$ even; not guaranteed for all perfect squares, so it need not be an integer.
- (C) $\(\frac{\sqrt{q}}{2}=\frac{k}{2}\)$, which is an integer only if $k$ is even; not always.
- (D) $\(q^2+q+\sqrt{q}=k^4+k^2+k\)$, clearly an integer for any integer $k$.
Thus the expressions that are always integers are (A) and (D).
- (A) $\(q^2+\sqrt{q}+1=k^4+k+1\)$, which is an integer for any integer $k$.
- (B) $\(q^3+\frac{q}{2}+3=k^6+\frac{k^2}{2}+3\)$. This is an integer only if $\(k^2\)$ is even, i.e., $k$ even; not guaranteed for all perfect squares, so it need not be an integer.
- (C) $\(\frac{\sqrt{q}}{2}=\frac{k}{2}\)$, which is an integer only if $k$ is even; not always.
- (D) $\(q^2+q+\sqrt{q}=k^4+k^2+k\)$, clearly an integer for any integer $k$.
Thus the expressions that are always integers are (A) and (D).
gmatclubot
Re: The variable $q$ is an integer and a perfect square. [#permalink]
11 Dec 2025, 02:28
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