jmc

Since $105=3\cdot 5\cdot 7$, by the Chinese Remainder Theorem, it suffices to find the remainders when $x$ is divided by $3$, $5$, and $7$. As $3+x$ leaves a remainder of $4$ when divided by $27=3^3$, it follows that $3+x\equiv 4(\mathrm{mod}3)$, and thus that $x\equiv 1(\mathrm{mod}3)$. Similarly, $$\begin{array}{rl}x& \equiv 9\equiv 4(\mathrm{mod}5)\\ x& \equiv 25\equiv 4(\mathrm{mod}7).\end{array}$$As $4\equiv 1(\mathrm{mod}3)$, it follows from the Chinese Remainder Theorem that $x\equiv \boxed{4}(\mathrm{mod}105)$.