For any natural numbers $m, n$, and $k$, such that $k$ divides both $m+2 n$ and $3 m$ $+4 n, k$ must be a common divisor of

Given: $k$ divides both $\boxed{m + 2n}$ and $\boxed{3m + 4n}$.

Our goal is to find what $k$ must be a common divisor of.

Easy Approach:

Take values!

Let $m=2, n=3$ (don't take 1 in questions like these).

$k$ can divide:

I) $3m+4n = 6+12 = \fbox{18}$

II) $m+2n = 2 + 6 = \fbox{8}$

Only $k=2$ can divide the above two numbers.

Now, let's check the options:

A) $2 m,3 n = {4,9}$ -> both are not divisible by $2$.

B) $2 m, n = {4,3}$ -> both are not divisible by $2$.

C) $m, n = {2,3}$ -> both are not divisible by $2$.

D) $m, 2n = {2,6}$ -> ✅ both divisible by $2$.

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Conceptual Approach:

If a number $k$ divides two expressions, then $k$ also divides any combination you can make by adding, subtracting, or multiplying these expressions by constants.

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Since $k$ divides both $(m + 2n)$ and $(3m + 4n)$, let's create a combination that isolates $m$.

From $(m + 2n)$, multiply by 2:

$k$ divides $2(m + 2n) = 2m + 4n$

Now we have:

$k$ divides $(2m + 4n)$

$k$ divides $(3m + 4n)$

Since $k$ divides both expressions, $k$ must also divide their difference:

$(3m + 4n) - (2m + 4n)$

$= 3m + 4n - 2m - 4n$

$= m$

Therefore, $k$ divides $m$.

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Now let's create a combination that isolates the $n$ terms.

From $(m + 2n)$, multiply by 3:

$k$ divides $3(m + 2n) = 3m + 6n$

Now we have:

$k$ divides $(3m + 6n)$

$k$ divides $(3m + 4n)$

Taking the difference:

$(3m + 6n) - (3m + 4n)$

$= 3m + 6n - 3m - 4n$

$= 2n$

Therefore, $k$ divides $2n$.

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$k$ must be a common divisor of $m$ and $2n$

This method uses the fundamental property that divisors are preserved under linear combinations. By cleverly choosing multiples and taking differences, we can isolate the terms we want to analyze. This technique is extremely useful for divisibility problems involving multiple variables.