$\begin{cases}x\equiv 1 \pmod{3}\\ x\equiv 2 \pmod{5}\\ x\equiv 3 \pmod{7}\\ x\equiv 4 \pmod{9}\\ x\equiv 5 \pmod{11}\end{cases}$
I am supposed to solve the system using the Chinese remainder theorem but $(3,5,7,9,11)\neq 1$ How can I transform the system so that I will be able to use the theorem?
Observe that $\displaystyle x\equiv4\pmod 9\implies x\equiv4\pmod3\equiv1$
Now, we can safely apply C.R.T on $$\begin{cases} x\equiv 2 \pmod5\\ x\equiv 3 \pmod7\\ x\equiv 4 \pmod9\\ x\equiv 5 \pmod{11}\end{cases}$$ as $5,7,9,11$ are pairwise relatively prime