[Combinatorics]
Problem: The mean, median, and unique mode of a list of positive integers are three consecutive integers in some order. Compute the least possible sum of the integers in the original list.

Solution: One possible list is $1,1,3,7$, which has mode 1 , median 2 , and mean 3 . The sum is $1+1+3+7=12$. A list with fewer than four numbers cannot produce a median and unique mode that are distinct from each other. To see this, first note that a list with one number has the same median and mode. In a list with two numbers, the mode is not unique if the numbers are different, and if the numbers are the same, the median and mode are equal. In a list of three numbers with a unique mode, the mode must occur twice. Hence the



mode is equal to the middle number of the three, which is the median. Thus a list with a median and unique mode that are different from each other must contain at least four numbers.

Now suppose that a list satisfying the given conditions sums to less than 12 . The mean must be greater than 1, and because the list contains at least four numbers, the mean must be exactly 2 . The median must also be greater than 1 , and if the mode is 4 , then the sum must be greater than 12 . Thus it remains to determine if a mean of 2 with mode 1 and median 3 can be achieved with a list of four or five positive integers. However, having two 1s in the list and a median of 3 forces the remaining numbers in each case to have a sum too large for a mean of 2 . The least possible sum is therefore $\mathbf{1 2}$.

Final answer: 12