Correct Answer: C\\ \\ Let $N$ be the product of $k$ consecutive positive integers. So, $N = n(n+1)(n+2)...(n+k-1)$, where $n$ is a positive integer.\\ \\ We know that the product of any $k$ consecutive positive integers is always divisible by $k!$. That is, $\frac{N}{k!}$ is an integer for any positive integer $n$ and $k$. This implies $k!$ is a factor of $N$.\\ \\ We need to find the minimum value of $k$ such that $N$ is always divisible by $720$. For $N$ to be *always* divisible by $720$ irrespective of the starting integer $n$, it must be that $k!$ itself is divisible by $720$.\\ \\ **Step 1: Find the prime factorization of $720$.**\\ $720 = 72 \times 10$ $72 = 8 \times 9 = 2^3 \times 3^2$ $10 = 2 \times 5$ Therefore, $720 = (2^3 \times 3^2) \times (2 \times 5) = 2^{3+1} \times 3^2 \times 5^1 = 2^4 \times 3^2 \times 5^1$.\\ \\ **Step 2: Determine the minimum $k$ such that $k!$ contains all prime factors of $720$.**\\ We need $k!$ to be divisible by $2^4$, $3^2$, and $5^1$. Let's examine the factorials for increasing values of $k$:\\ \begin{itemize} \item **For $k=4$:** $4! = 4 \times 3 \times 2 \times 1 = 24 = 2^3 \times 3^1$. This does not contain $2^4$, $3^2$, or $5^1$. Thus, $4!$ is not divisible by $720$. For example, $N = 1 \times 2 \times 3 \times 4 = 24$, which is not divisible by $720$. \item **For $k=5$:** $5! = 5 \times 4! = 5 \times (2^3 \times 3^1) = 2^3 \times 3^1 \times 5^1 = 120$. This contains $5^1$, but still lacks $2^4$ and $3^2$. Thus, $5!$ is not divisible by $720$. For example, $N = 1 \times 2 \times 3 \times 4 \times 5 = 120$, which is not divisible by $720$. \item **For $k=6$:** $6! = 6 \times 5! = (2 \times 3) \times (2^3 \times 3^1 \times 5^1) = 2^{1+3} \times 3^{1+1} \times 5^1 = 2^4 \times 3^2 \times 5^1 = 720$. This contains all the required prime factors ($2^4$, $3^2$, $5^1$). Thus, $6!$ is divisible by $720$.\\ \end{itemize} Since $6!$ is divisible by $720$, the product of any $6$ consecutive positive integers will always be divisible by $720$. For any $k < 6$, $k!$ is not divisible by $720$, meaning there exist products of $k$ consecutive integers that are not divisible by $720$.\\ \\ Therefore, the minimum value of $k$ is $6$.\\ \\ The final answer is $\boxed{C}$