Final answer:
The highest possible average of set S occurs when values are maximized under the given constraints. This involves setting the three values above the median M to 2M and the three below to M-1, M-2, and M-3. The highest average is thus (9M - 6)/7, with M being the largest integer that satisfies the set's conditions.
Step-by-step explanation:
The question asks us to find the highest possible average of a set S containing seven distinct integers with the median being integer M and all values being equal to or less than 2M. The maximum average will occur when all values are as large as possible given the constraints.
Since the median of set S is M, we know that there are three values below M and three above M in the set. To maximize the average, we should make the three values above M as large as possible, which would be 2M (since no value can exceed 2M). As for the three integers below M, they should be as close to M as possible to obtain the highest sum, ideally, these would be one less than M, two less than M, and three less than M.
The set S would look like this for the maximum sum: {M-3, M-2, M-1, M, 2M, 2M, 2M}. Calculating the sum, we have 3M - 6 + 6M, which simplifies to 9M - 6. To find the average, we divide this by 7, the number of integers in set S, resulting in (9M - 6)/7. Since 9M is divisible by 7 but -6 is not, M must also be chosen in such a way that when 6 is subtracted, the result is still divisible by 7 to yield an integer average. Hence, if M is a multiple of 7, the average will be maximized.
The highest average is reached when set S comprises the integers as close to M and 2M as the constraints allow. Therefore, the optimal average (arithmetic mean) for set S would be (9M - 6)/7, where M is the maximum integer such that all members of set S are integers and no larger than 2M.