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

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

cdquestions Admin · Accepted Answer

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  $2n$

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Answer

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

The Correct Option is D

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