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

Accepted Answer

B. Given that $k$ divides $(m + 2n)$ and $k$ divides $(3m + 4n)$. Since $k \mid (m + 2n)$, $k$ must also divide $2(m + 2n) = 2m + 4n$. Now, $k$ divides both $(3m + 4n)$ and $(2m + 4n)$, so it must divide their difference: $$k \mid [(3m + 4n) - (2m + 4n)] \implies k \mid m$$ Since $k \mid m$ and $k \mid (m + 2n)$, $k$ must divide $(m + 2n) - m = 2n$. Therefore, $k$ must be a common divisor of $m$ and $2n$.

Common divisor of algebraic expressions

Explanation

Practise this under exam conditions