The mean, median, unique mode, and range of a collection of eight integers are all equal to 8. The largest integer that can be an element of this collection is 
$\text{(A) }11 \qquad \text{(B) }12 \qquad \text{(C) }13 \qquad \text{(D) }14 \qquad \text{(E) }15$
As the unique mode is $8$, there are at least two $8$s.
As the range is $8$ and one of the numbers is $8$, the largest one can be at most $16$.
If the largest one is $16$, then the smallest one is $8$, and thus the mean is strictly larger than $8$, which is a contradiction.
If we have 2 8's we can add find the numbers 4, 6, 7, 8, 8, 9, 10, 12.
This is a possible solution but has not reached the maximum.
If we have 4 8's we can find the numbers 6, 6, 6, 8, 8, 8, 8, 14.
We can also see that they satisfy the need for the mode, median, and range to be 8. This means that the answer will be
$\boxed{\text{(D)}\ 14 }$. ~By QWERTYUIOPASDFGHJKLZXCVBNM