smc

Consider $3$ consecutive integers $a,b,$ and $c$. Exactly one of these integers must be divisible by 3; WLOG, suppose $a$ is divisible by 3. Then $a\equiv 0(\mathrm{mod}3),b\equiv 1(\mathrm{mod}3),$ and $c\equiv 2(\mathrm{mod}3)$. Squaring, we have that $a^2\equiv 0(\mathrm{mod}3),b^2\equiv 1(\mathrm{mod}3),$ and $c^2\equiv 1(\mathrm{mod}3)$, so $a^2+b^2+c^2\equiv 2(\mathrm{mod}3)$. Therefore, no member of $S$ is divisible by 3. Now consider $3$ more consecutive integers $a,b,$ and $c$, which we will consider mod 11. We will assign $k$ such that $a\equiv k(\mathrm{mod}11),b\equiv k+1(\mathrm{mod}11),$ and $c\equiv k+2(\mathrm{mod}11)$. Some experimentation shows that when $k=4,a\equiv 4(\mathrm{mod}11)$ so $a^2\equiv 5(\mathrm{mod}11)$. Similarly, $b\equiv 5(\mathrm{mod}11)$ so $b^2\equiv 3(\mathrm{mod}11)$, and $c\equiv 6(\mathrm{mod}11)$ so $c^2\equiv 3(\mathrm{mod}11)$. Therefore, $a^2+b^2+c^2\equiv 0(\mathrm{mod}11)$, so there is at least one member of $S$ which is divisible by 11. Thus, $\boxed{\text{No member of S is divisible by}3\text{ but some member is divisible by }11}$ is correct.