All of the even numbers from 2 to 98 inclusive, excluding those ending in 0, are multiplied together.  What is the rightmost digit (the units digit) of the product?
$\text{(A)}\ 0 \qquad \text{(B)}\ 2 \qquad \text{(C)}\ 4 \qquad \text{(D)}\ 6 \qquad \text{(E)}\ 8$
All the tens digits of the product will be irrelevant to finding the units digit.  Thus, we are searching for the units digit of $(2\cdot 4\cdot 6 \cdot 8) \cdot (2 \cdot 4 \cdot 6 \cdot 8) \cdot (2\cdot 4\cdot 6 \cdot 8) \cdot ...$
There will be $10$ groups of $4$ numbers.  The number now can be rewritten as $(2\cdot 4 \cdot 6 \cdot 8)^{10}$.
Multiplying, we get $384^{10}$.
Again, we can disregard the tens and hundreds digit of $384$, since we only want the units digit of the number, leaving $4^{10}$.
Now, we try to find a pattern to the units digit of $4^n$.  To compute this quickly, we once again discard all tens digits and higher.
$4^1 = 4$.
$4^2 = 4\cdot 4 = 1\underline{6}$, discard the $1$.
$4^3 = 1 \cdot 4 = \underline{4}$
$4^4 = 4 \cdot 4 = 1\underline{6}$, discard the $1$.
$4^5 = 1 \cdot 4 = \underline{4}$
Those equalities are, in reality, congruences $\mod {10}$.
Thus, the pattern of the units digits is $\{4, 6, 4, 6, 4, 6, 4, 6, 4, 6\}$.  The cycle repeats so that term $n$ is the same as term $n+2$.  The tenth number in the cycle is $6$, giving an answer of $\boxed{D}$.