I ran into this version of the Chinese Remainder Theorem:
Show that the solutions to the simultaneous system of congruences $$x\equiv a_1\pmod{m_1},$$ $$x\equiv a_2\pmod{m_2},$$ $$...$$ $$x\equiv a_r\pmod{m_r},$$ where the $m_j$ are pairwise relatively prime, are given by $$x\equiv a_1M_1^{\phi(m_1)}+a_2M_2^{\phi(m_2)}+\cdots+a_rM_r^{\phi(m_r)}\pmod{M},$$ where $M=m_1m_2\cdots m_r$ and $M_j=M/m_j$ for $j=1, 2, ..., r.$
This is basically the exact Chinese Remainder Theorem, except that instead of $x_jM_jy_j$, we have $x_jM_j^{\phi(m_j)}$. Would it then suffice to show that $x_jM_j^{\phi(m_j)}\equiv x_jM_jy_j\pmod{M}$? If so, how could I go about it? Thanks in advance!