2014 AMC 10B Problem 18

The total is $11 \cdot 10 = 110$. Sort the list $a_1 \le \cdots \le a_{11}$; the median gives $a_6 = 9$. To make $a_{11}$ as large as possible, make $a_1 + \cdots + a_{10}$ as small as possible.

Say $8$ appears $k$ times. All the 8s sit below the median, so $k \le 5$, and every other value appears at most $k - 1$ times. The value $9$ already appears once (as $a_6$), and each entry above the median is at least $9$.

• $k = 5$: $a_1..a_5 = 8$; then $9$ at most four times, so $a_7, a_8, a_9 = 9$, $a_{10} = 10$. Sum of ten: $40 + 36 + 10 = 86$, so $a_{11} = 24$.
• $k = 4$: bottom $1, 8, 8, 8, 8$; $a_7, a_8 = 9$, $a_9, a_{10} = 10, 10$. Sum $33 + 27 + 20 = 80$, so $a_{11} = 30$.
• $k = 3$: bottom $1, 1, 8, 8, 8$; $a_7 = 9$, $a_8, a_9 = 10, 10$, $a_{10} = 11$. Sum $26 + 18 + 20 + 11 = 75$, so $a_{11} = 35$.
• $k = 2$: bottom $1, 2, 3, 8, 8$; $a_7..a_{10} = 10, 11, 12, 13$. Sum $22 + 9 + 46 = 77$, so $a_{11} = 33$.

The best is $k = 3$: the list $1, 1, 8, 8, 8, 9, 9, 10, 10, 11, 35$ has sum $110$, median $9$, and unique mode $8$.

The answer is $\boxed{\textbf{(E)}\ 35}$.

A fixed mean turns "maximize one entry" into "minimize the rest." The mode condition cuts both ways: more 8s let you repeat 9s and 10s above the median, but they also push out cheap entries like 1 below it. Trying each frequency of the mode is a short, exhaustive search, and the answer choices are exactly the values these cases produce.

• Using five 8s to fill the bottom, which forces the other values up and yields 24; fewer 8s lets 1s in and gives a bigger maximum.
• Placing $9$ three times when $8$ appears only three times, breaking the uniqueness of the mode.