Discussion

Explanation:

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

k divides m + 2n ⇒ m + 2n = a × k   ...(1)
k divides 3m + 4n ⇒ 3m + 4n = b × k   ...(2)

(1) × 3 - (2)
⇒ 3m + 6n - (3m + 4n) = 3ak - bk
⇒ 2n = (3a - b)k
∴ 2n is divisible by k

(2) - (1) × 2
⇒ 3m + 4n - 2(m + 2n) = bk - 2ak
⇒ m = (b - 2a)k
∴ m is divisible by k

∴ 2n and m are divisible by k
⇒ k is common divisor of 2n and m.

Hence, option (b).

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