6. From 10 different natural numbers whose sum is 55, after removing 3 numbers, the sum of the remaining numbers is $\frac{7}{11}$ of 55, then the maximum product of the three removed numbers is ( )
A. 280
B. 270
C. 252
D. 216
6. Solution: The sum of the remaining numbers is: $55 \times \frac{7}{11}=35$, the sum of the numbers taken out is: $55-35=20$. To maximize the product of the three numbers taken out, the numbers should be as close as possible. We know that $6+7+8=21$, so we take $5 \times 7 \times 8=280$

Answer: A