P = {6, 3, 0, d, 4, 14, 9, 2d} If d is the smallest

d = smallest positive integer 'm' which the range of remainder when $$\frac{multiple\ of\ 3}{d}=3$$
Question: By what percent is the median of the numbers in P smaller than the mean of the numbers in P?
The only way in which a multiple of 3 can be divided, and give a remainder that is equal to 3 is when the multiple of 3 is divided by a number greater than 3.
Therefore, minimum possible value of d = 3+1 = 4
Considering 15 as a multiple of 3
=> 15/4 = 3 remainder 3
So, 'd' can be said to be equal to 4.
Therefore, p = {6, 3, 0, 4, 4, 14, 9, 8}
When arranged in ascending order =>
p = {0, 3, 4, 4, 6, 8, 9, 14}
Median of p = (4+6)/2 = 10/2 = 5
$$Mean=\frac{sum\ of\ fx}{sum\ of\ f}=\frac{0+3+4+4+6+8+9+14}{8}$$
$$=\frac{48}{8}=6$$
% by which medium is less than mean =
$$\frac{6-5}{6}\cdot\frac{100}{1}=\frac{100}{6}=16.7\%$$
Answer = option C