If 15 × 14 × 13 × ... × 3 × 2 × 1 = $3^m\times n$ where m and n are positive integers, then what is the maximum value of m?

The correct option is 6.

To find the maximum value of $m$ such that $3^m$ divides the product $15\times 14\times 13\times ...\times 2\times 1$, we need to determine the exponent of the prime number 3 in the prime factorization of this product.

The given product is the factorial of 15, denoted as 15!:
$15!=15\times 14\times 13\times ...\times 3\times 2\times 1$

To find the total number of times the prime factor 3 appears in 15!, we identify all the multiples of 3 that are less than or equal to 15. These numbers are:
3, 6, 9, 12, and 15.

Now, let's find the number of times 3 divides each of these multiples:
- $3=3^1$ (contributes one 3)
- $6=2\times 3^1$ (contributes one 3)
- $9=3^2$ (contributes two 3s)
- $12=2^2\times 3^1$ (contributes one 3)
- $15=5��3^1$ (contributes one 3)

Summing up the contributions of the factor 3, we get:
$1+1+2+1+1=6$

Alternatively, we can use Legendre's formula to find the exponent of a prime $p$ in $n!$, which is given by:
$E\left(p\right)=\lfloor \frac{n}{p}\rfloor +\lfloor \frac{n}{p^2}\rfloor +\lfloor \frac{n}{p^3}\rfloor +...$
where $\lfloor x\rfloor$ represents the greatest integer less than or equal to $x$.

Substituting $n=15$ and $p=3$:
$E\left(3\right)=\lfloor \frac{15}{3}\rfloor +\lfloor \frac{15}{9}\rfloor +\lfloor \frac{15}{27}\rfloor +...$
$E\left(3\right)=5+1+0=6$

Therefore, the prime 3 appears exactly 6 times in the prime factorization of 15!. Thus, the maximum value of the integer exponent $m$ is 6.