The product $8 \times 48 \times 81$ is divisible by $6^{k}$. The largest possible integer value of $k$ is
(A) 7
(B) 6
(C) 5
(D) 4
(E) 3
We note that $8 \times 48 \times 81=2^{3} \times\left(2^{4} \times 3\right) \times 3^{4}=2^{7} \times 3^{5}=2^{2} \times 2^{5} \times 3^{5}=2^{2} \times(2 \times 3)^{5}=2^{2} \times 6^{5}$. After $6^{5}$ is divided out from $8 \times 48 \times 81$, the quotient has no factors of 3 and so no further factors of 6 can be divided out.

Therefore, the largest integer $k$ for which $6^{k}$ is a divisor of $8 \times 48 \times 81$ is $k=5$.

ANswer: (C)