Find all solutions, if any, to the system of congruences
$$\begin{align} x&\equiv 7 \pmod{9}\\ x&\equiv 4 \pmod{12}\\ x&\equiv 16 \pmod{21} \end{align}$$
Solution: Using the Chinese Remainder theorem, we get that this system is equivalent to the 5 equations: $$\begin{align} x&\equiv 7 \pmod{9} \\ x&\equiv 0 \pmod{4} \\ x&\equiv 1 \pmod{3} \\ x&\equiv 2 \pmod{7} \\ x&\equiv 1 \pmod{3} \\ \end{align}$$
The 3rd and 5th equations are superfluous, and the total system has general solution $x\equiv16 \pmod {252}$."
I can't seem to get 16 all I get is 64, why?
I do it like this $$\begin{align} x&\equiv a_1M_1y_1+a_2M_2y_2+a_3M_3y_3 \pmod{16}\\ &\equiv 7\cdot28\cdot4 + 0 + 2\cdot36\cdot4 \pmod{16}\\ &\equiv 1072\pmod{16} \\ &\equiv 64 \pmod{16} \end{align}$$