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Plugging In: Basic Technique

If $$k$$ and $$m$$ are positive integers, and $$k=4m$$, then which of the following must be true?
Incorrect. [[Snippet]] 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. [[Snippet]] 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. [[Snippet]] 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. [[Snippet]] 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. [[Snippet]] 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.