Hi Heureka,
I have been learning from some of your modular arithemtic answers. I thank you for providing them. 
I looked at this one and thought there should be an easier method.
This is what I came up with. Perhaps you could take a look and if you find fault with my logic then let me know.
\(\begin{align*} 3+x &\equiv 2^2 \pmod{3^3} \\ 5+x &\equiv 3^2 \pmod{5^3} \\ 7+x &\equiv 5^2 \pmod{7^3} \end{align*}\)
\(3+x\equiv 4 \quad \pmod{3^3}\\ x\equiv 1 \quad \pmod{3^3}\\ so\\ x\equiv 1 \quad \pmod{3}\\ x=1+3m\qquad \qquad (1) \)
\(5+x\equiv 9 \quad \pmod{5^3}\\ x\equiv 4 \quad \pmod{5^3}\\ x\equiv 4 \quad \pmod{5}\\ x\equiv -1 \quad \pmod{5}\\ x=-1+5n\qquad \qquad (2)\)
\(7+x\equiv 25 \quad \pmod{7^3}\\ x\equiv 18 \quad \pmod{7^3}\\ x\equiv 18 \quad \pmod{7}\\ x\equiv 4 \quad \pmod{7}\\ x=4+7k\qquad \qquad (3)\)
Solve the first 2 simultaneously
\(1+3m=-1+5n\\ 2=5n-3m\\ \text{One obvious solution for this is n=1 and m=1}\\ 2=5(1)-3(1)\\ \text{now I can add 15f and take it away again to get the general diophantine solution.}\\ 2=5(1+3f)-3(1+5f)\\ n=1+3f\\ x=-1+5n\\ x=-1+5(1+3f)\\ x=4+15f\\ \)
I could work it out with m instead of n but x in terms of f will be the same.
So now I have
\(x=4+7k \quad (3) \qquad and \qquad x=4+15f\\ \text{since }\\x\equiv 4 \pmod{7}\qquad and\\ x\equiv4\pmod{15}\\ \text{It follows that}\\ x\equiv 4 \pmod{7*15}\\ x\equiv 4 \pmod{105}\\ \)
So when x is divided by 105 the remainder is 4
+118609 
