smc

Note that for all polynomials $f(x)$, $f(x+6)\equiv f(x)(\mathrm{mod}6)$. Proof: If $f(x)=a_n{x}^n+a_{n-1}{x}^{n-1}+...+a_0$, 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 $(\mathrm{mod}6)$. This leaves $a_n{x}^n+a_{n-1}{x}^{n-1}+...+a_0$, which is $f(x)$, so $f(x+6)\equiv f(x)(\mathrm{mod}6)$. 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 $\boxed{17}$