CRT Difficult

Problem - 2662

Solve $$\left\{ \begin{array}{rcl} 4x & \equiv 14 &\pmod{15}\\ 9x & \equiv 11 &\pmod{20}\\ \end{array}\right.$$


Because $gcm(15,20)\ne 1$, CRT cannot be applied directly. As such, we decompose the relation first. The given equations are equivalent to: $$\left\{ \begin{array}{rcl} 4x & \equiv 14 &\pmod{3}\\ 4x & \equiv 14 &\pmod{5}\\ 9x & \equiv 11 &\pmod{4}\\ 9x & \equiv 11 &\pmod{5} \end{array} \right. $$


This system can be simplified as $$\left\{ \begin{array}{rcl} x & \equiv 2 &\pmod{3}\\ x & \equiv -1 &\pmod{5}\\ x & \equiv 3 &\pmod{4}\\ x & \equiv 1 &\pmod{5} \end{array} \right. $$


The $2^{st}$ and the $4^{th}$ equations contradict to each other. Hence, this system does not have any solution.

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