Question 1109303
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In a set of 11 positive integers, the 6th one is the median, which is 9:
x, x, x, x, x, 9, x, x, x, x, x<br>
The unique mode is 8; let's use only two 8's and see what the largest integer in the set can be:
x, x, x, 8, 8, 9, x, x, x, x, x<br>
The mean is 10, so the sum of all 11 numbers is 110.  We are trying to find the largest possible value for a number in the set; that means we want all the other numbers to be as small as possible.  Since 8 is the unique mode and we are trying to use only two of them, the other numbers must all be different; for the sum of all 11 numbers to be 110, we get this:
1, 2, 3, 8, 8, 9, 10, 11, 12, 13, 33<br>
Now let's use three 8's; that will allow us to use some smaller numbers twice, making it possible that we might get a larger number than 33 in the set.  Using the smallest possible numbers, keeping our median of 9 and unique mode of 8, now using three 8s, we get this:
1, 1, 8, 8, 8, 9, 9, 10, 10, 11, 35<br>
Yes, using three 8's for the unique mode, we were able to get 35 for the largest number in the set.<br>
What about four 8's for the unique mode?  Using the same logic, we get this:
1, 8, 8, 8, 8, 9, 9, 9, 10, 10, 30.<br>
With the fourth 8, the largest number we can get in the set is 30.<br>
So the final answer is...<br>
The largest possible number in the set is 35.