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Is n an integer?

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Is n an integer?

by Vincen » Wed Feb 21, 2018 2:07 am
Is n an integer? $$(1)\ \ \ 3n^2\ is\ an\ integer.$$ $$(2)\ \ \frac{\sqrt{n}}{3}\ is\ an\ \ integer.$$ Experts, is not sufficient the statement (1)? Why not? Also, how can I use the statement (2) to get an answer?

Can you clarify this for me?

Thanks in advanced.
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Source: — Data Sufficiency |

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by ErikaPrepScholar » Wed Feb 21, 2018 8:31 am
Statement 1
Let's try an example where 3n^2 is an integer: we'll let $$3n^2=1$$ Solving for n gives $$n^2=\frac{1}{3}$$ $$n=\frac{1}{\sqrt{3}}$$ This is not an integer. Insufficient.

Statment 2
We know $$\frac{\sqrt{n}}{3}=integer$$ Multiplying an integer by another integer will always give an integer, so if we multiply both sides by 3 $$\sqrt{n}=3\left(integer\right)$$ $$\sqrt{n}=integer$$ Squaring an integer is that same as multiplying that integer by itself. So again, we're multiplying an integer by an integer, which will always give an integer. So if we square both sides $$n=integer^2$$ $$n=integer$$ So n will always be an integer. Sufficient.
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