AMC 12 Mock Exam 1 Problem 2

Write the list in order as $a\le b\le c\le d\le e$. The median is $c=12$ and the sum is $5\cdot14=70$.

The mode $9$ must occur at least twice (otherwise every value occurs once and there is no mode). Since $d,e\ge c=12>9$ and $c=12\ne 9$, both $9$'s are among $a,b$; hence $a=b=9$, and $d+e=70-9-9-12=40$.

If $d=12$, then $12$ occurs twice (as $c$ and $d$), tying with $9$, so the mode would not be unique. Thus $d\ge13$ and $e=40-d\le 27$. The list $9,9,12,13,27$ has mean $14$, median $12$, and unique mode $9$. The answer is $\boxed{27}$.

(The tempting value $28$ would need $d=12$, which creates a second mode.)