6. The product of two natural numbers is 2160. Their Greatest Common Divisor is odd and greater than 1. Denoting by $m$ the least common multiple of the two numbers, we have:
(A) $m<200$
(D) $600<m<800$
(B) $200<m<400$
(C) $400<m<600$
(6) The correct answer is $(D)$.

Let's factorize: $2160=2^{4} \cdot 3^{3} \cdot 5$. Considering that the product of the GCD and LCM must still be equal to $2160$ and that, moreover, the GCD is always a divisor of the LCM, we conclude that the GCD must be 3.

Indeed, being odd and greater than 1, it must be either 5 or $3^{k}$ (with $1 \leq k \leq 3$), or $5 \cdot 3^{k}$ (with $1 \leq k \leq 3$). However, if 5 were to divide the GCD, then 5 would also divide the LCM, and thus 25 should be a divisor of 2160, which is absurd. Similarly, if $3^{2}$ were to divide the GCD, then $3^{4}$ would divide 2160, which is also absurd. Therefore, the GCD of the two numbers is indeed 3.

The LCM will then be $2160: 3=720$.

Question proposed by Matteo Protopapa.