A list of five positive integers has [mean](https://artofproblemsolving.com/wiki/index.php/Mean) $12$ and [range](https://artofproblemsolving.com/wiki/index.php/Range) $18$. The [mode](https://artofproblemsolving.com/wiki/index.php/Mode) and [median](https://artofproblemsolving.com/wiki/index.php/Median) are both $8$. How many different values are possible for the second largest element of the list? 
$\mathrm{(A) \ 4 } \qquad \mathrm{(B) \ 6 } \qquad \mathrm{(C) \ 8 } \qquad \mathrm{(D) \ 10 } \qquad \mathrm{(E) \ 12 }$
Let $a$ be the smallest element, so $a+18$ is the largest element. Since the mode is $8$, at least two of the five numbers must be $8$. The last number we denote as $b$.
Then their average is $\frac{a + (8) + (8) + b + (a+18)}5 = 12 \Longrightarrow 2a + b = 26$. Clearly $a \le 8$. Also we have $b \le a + 18 \Longrightarrow 26-2a \le a + 18 \Longrightarrow 8/3 < 3 \le a$. Thus there are a maximum of $6$ values of $a$ which corresponds to $6$ values of $b$; listing shows that all such values work. The answer is $\boxed{B}$.