The sum of $5$ consecutive even integers is $4$ less than the sum of the first $8$ consecutive odd counting numbers. What is the smallest of the even integers?
$\textbf{(A) } 6 \qquad\textbf{(B) } 8 \qquad\textbf{(C) } 10 \qquad\textbf{(D) } 12 \qquad\textbf{(E) } 14$
It is a well-known fact that the sum of the first $n$ odd numbers is $n^2$.  This makes the sum of the first $8$ odd numbers equal to $64$.
Let $n$ be equal to the smallest of the $5$ even integers.  Then $(n+2)$ is the next highest, $(n+4)$ even higher, and so on.
This sets up the equation $n+(n+2)+(n+4)+(n+6)+(n+8)+4=64$
Now we solve:
\[5n+24=64\]
\[5n=40\]
\[n=8\]
Thus, the smallest integer is $\boxed{\textbf{(B) } 8}$.