Question

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

• A. m and n
• B. 2m and 3n
• C. 2m and n
• D. m and 2n

Answer 1

Answer: (D)

m and 2n

Answer 2

Given that,
$k$ divides $m+2n$ and $3m+4n$.
Since $k$ divides $(m+2n)$
$⇒ k$ will also divide $3(m+2n)$
$⇒ k$ divides $3m+6n$
Similarly,
$k$ divides $3m+4n$.
We know that, if two numbers $a$ and $b$ both are divisible by $c$, then their $(a-b)$ is also divisible by $c$.
By the same phenomenon,
$[(3m+6n)-(3m+4n)]$ is divisible by $k$.
So, $2n$ is also divisible by $k$.
Now, $(m+2n)$ is divisible by $k$,
$⇒ 2(m+2n)$ is also divisible by $k$.
$⇒ 2m+4n$ is also divisible by k.
So, $[(3m+4n)-(2m+4n)] = m$ is also divisible by $k$.
Therefore, $m$ and $2n$ are also divisible by $k$.

So, the correct option is (D): m and $2n$