Problem - 2639
Solve $$\left\{ \begin{array}{rcl} x &\equiv 2 &\pmod{3}\\ x &\equiv 2 &\pmod{5}\\ x &\equiv -3 &\pmod{7}\\x &\equiv -2 &\pmod{13} \end{array}\right.$$
This system can be solved by the Chinese Remainder Theorem (CRT). $$m=3\times 5\times 7\times 13 = 1365$$
and, $$M_1 = 455, M_2 = 273, M_3 = 195, M_4=105$$
accordingly, $$M_1^{-1} = -1, M_2^{-1} = 2, M_3^{-1} = -1,M_4^{-1} = -1$$
Therefore, $$\begin{array}{rl} x=&455\times (-1) \times 2 +273 \times 2 \times 2 + 195 \times (-1) \times (-3) +\\ & 105\times (-1) \times (-2)\\ =&\boxed{557\pmod{1365}}\end{array}$$