Final answer:
To find the largest possible number, we calculate the range using the constraints given and find that the largest number can be 24 while ensuring all six integers are different and add up to 66, with one being the mean and another the range.
Step-by-step explanation:
The student's question involves finding the largest possible number in a set of six different integers that add up to 66, considering that one of the numbers is the mean of the set and another is the range of the set.
We know that the sum of the six integers is 66, so the mean is 66 ÷ 6, which is 11. Let's assume the smallest number in the set is x. Because one of the numbers is also the range, we can express the largest number as x + range. Since the mean is 11, we have five other numbers to consider, and to maximize the largest number, we need to minimize the other five.
Let's say the numbers are x, (x + 1), (x + 2), 11 (the mean), (x + range - 1), and (x + range). To minimize, we make x as low as possible, so we start from 1 (since they must be different integers). Therefore, we have 1, 2, 3, 11, x + range - 1, and x + range as our six numbers. The sum of the first four is 17, leaving 49 for the last two numbers.
The range would be (x + range) - 1, so we can calculate the largest possible value for the range. The equation to solve is:
2(x + range) - 1 = 49
Two times the largest number minus 1 should be equal to 49.
2*range + 1 = 49
2*range = 48
range = 24.
Lastly, the largest number would be 1 (the smallest number) + 24 (the range), which equals 25. However, since they need to be six different integers, and we already have 11 in the set, the largest number we can have without repeating is 24. Thus, the correct answer is a) 24.