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

UPSC Quizroom · Accepted Answer

Step 1: Express a combination to eliminate n

Multiply the first equation by 2: 2(m + 2n) = 2m + 4n → k | (2m + 4n)

Subtract the second equation: (2m + 4n) − (3m + 4n) = −m → k | (−m) → k | m

So k divides m.

Step 2: Express a combination to eliminate m

Multiply first equation by 3: 3(m + 2n) = 3m + 6n → k | (3m + 6n)

Subtract second equation: (3m + 6n) − (3m + 4n) = 2n → k | 2n

So k divides 2n.

✅ Step 3: Conclusion

k is a common divisor of m and 2n.

Answer: m and 2n

🎯 UPSC Quiz Room