Let $f(x)=x^2+3x+2$ and let $S$ be the set of integers $\{0, 1, 2, \dots , 25 \}$. 
The number of members $s$ of $S$ such that $f(s)$ has remainder zero when divided by $6$ is:
$\textbf{(A)}\ 25\qquad \textbf{(B)}\ 22\qquad \textbf{(C)}\ 21\qquad \textbf{(D)}\ 18 \qquad \textbf{(E)}\ 17$
Note that for all polynomials $f(x)$, $f(x + 6) \equiv f(x) \pmod 6$.  
Proof:
If $f(x) = a_nx^n + a_{n-1}x^{n-1} + ... + a_0$, then $f(x+6) = a_n(x+6)^n + a_{n-1}(x+6)^{n-1} +...+ a_0$.  In the second equation, we can use the binomial expansion to expand every term, and then subtract off all terms that have a factor of $6^1$ or higher, since subtracting a multiple of $6$ will not change congruence $\pmod 6$.  This leaves $a_nx^n + a_{n-1}x^{n-1} + ... + a_0$, which is $f(x)$, so $f(x+6) \equiv f(x) \pmod 6$.
So, we only need to test when $f(x)$ has a remainder of $0$ for $0, 1, 2, 3, 4, 5$.  The set of numbers $6, 7, 8, 9, 10, 11$ will repeat remainders, as will all other sets.  The remainders are $2, 0, 0, 2, 0, 0$.
This means for $s=1, 2, 4, 5$, $f(s)$ is divisible by $6$.  Since $f(1)$ is divisible, so is $f(s)$ for $s=7, 13, 19, 25$, which is $5$ values of $s$ that work.  Since $f(2)$ is divisible, so is $f(s)$ for $s=8, 14, 20$, which is $4$ more values of $s$ that work.  The values of $s=4, 5$ will also generate $4$ solutions each, just like $f(2)$.  This is a total of $17$ values of $s$, for an answer of $\boxed{\textbf{(E)}}$