Plugging In: Basic Technique
If $$k$$ and $$m$$ are positive integers, and $$k=4m$$, then which of the following must be true?
Incorrect.
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Plug in $$m = 3$$:
>$$k = 4(3) = 12$$.
Eliminate this answer choice because $$\frac{m}{4(12)} = \frac{3}{48} = \frac{1}{16}$$ is _not_ an integer.
Correct.
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Plug in $$m = 3$$:
>$$k = 4(3) = 12$$.
Since $$\frac{k}{4} = \frac{12}{4}= 3$$ is a divisor of $$m$$ ($$m = 3$$), this answer choice is not eliminated. This is the correct answer because all other answer options are eliminated for the same plug-in.
Also, $$\frac{k}{4} = \frac{4m}{4} = m$$, which is a divisor of $$m$$.
Incorrect.
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Plug in $$m = 3$$:
>$$k = 4(3) = 12$$.
Eliminate this answer choice because $$\frac{k}{2} = \frac{12}{2} = 6$$ is _not_ a divisor of $$m$$ ($$m = 3$$).
Incorrect.
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Plug in $$m = 3$$:
>$$k = 4(3) = 12$$.
Therefore,
>$$2k = 2(12) = 24$$,
>$$2m = 2(3) = 6$$.
Eliminate this answer choice because $$2k$$ is _not_ a factor of $$2m$$.
Incorrect.
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Plug in $$m = 3$$:
>$$k = 4(3) = 12$$.
Therefore,
>$$4m = 4(3) = 12$$,
>$$\frac{k}{4} = \frac{12}{4} = 3$$.
Eliminate this answer choice because $$4m$$ is _not_ a divisor of $$\frac{k}{4}$$.
$$4m$$ is a divisor of $$\displaystyle \frac{k}{4}$$.
$$2k$$ is a factor of $$2m$$.
$$\displaystyle \frac{k}{2}$$ is a divisor of $$m$$.
$$\displaystyle \frac{k}{4}$$ is a divisor of $$m$$.
$$\displaystyle \frac{m}{4k}$$ is an integer.
