Question 3.
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
2m and 3n
B
m and 2n
C
2m and n
D
m and n
Question Explanation
Text ExplanationVideo Explanation
It is given that k divides m+2n and 3m+4n.
Since k divides (m+2n), it implies k will also divide 3(m+2n). Therefore, k divides 3m+6n.
Similarly, we know that k divides 3m+4n.
We know that if two numbers a, and b both are divisible by c, then their difference (a-b) is also divisible by c.
By the same logic, we can say that {(3m+6n)-(3m+4n)} is divisible by k. Hence, 2n is also divisible by k.
Now, (m+2n) is divisible by k, it implies 2(m+2n) =2m+4n is also divisible by k.
Hence, {(3m+4n)-(2m+4n)} = m is also divisible by k.
Therefore, m, and 2n are also divisible by k.
The correct option is B
