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

The Correct Option is D

Given that, \( k \) divides \( m + 2n \) and \( 3m + 4n \). 

Since \( k \) divides \( m + 2n \), 
\(\Rightarrow k\) also divides \( 3(m + 2n) \)  
\(\Rightarrow k \) divides \( 3m + 6n \). 

Also, \( k \) divides \( 3m + 4n \). 

Now, if \( k \) divides both \( 3m + 6n \) and \( 3m + 4n \), then it also divides their difference: 
\[ (3m + 6n) - (3m + 4n) = 2n \] So, \( k \) divides \( 2n \). 

Again, since \( k \) divides \( m + 2n \), \[ \Rightarrow k \text{ divides } 2(m + 2n) = 2m + 4n \] And \( k \) divides \( 3m + 4n \). 
Taking the difference: \[ (3m + 4n) - (2m + 4n) = m \] So, \( k \) divides \( m \). 

Hence, \( m \) and \( 2n \) are divisible by \( k \). 

Correct option: (D) \( m \) and \( 2n \).