Having A Problem With Chinese Remainder Theorem 
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
$$\begin{align}
x&\equiv a_1M_1y_1+a_2M_2y_2+a_3M_3y_3 \pmod{16}\\
&\equiv 7\cdot28\cdot4 + 0 + 2\cdot36\cdot4 \pmod{16}\\
&\equiv 1072\pmod{16} \\
&\equiv 64 \pmod{16}
\end{align}$$

I do small ones of these by hand.  From the first, we have $x=7+9k$  Taking that mod $4$, we get $x=1+3k \pmod 4$ and we want this to be $\equiv 0$.  By inspection, $k=1$ works and we have $x=16 \pmod {36}$.  Then $16 \equiv 2 \pmod 7$ and we are done, arriving at the solution $16 \pmod {252}$ 
You should be working mod $9 \cdot 4 \cdot 7=252,$ not mod $16$.  Note that $64 \equiv 0 \pmod {16}$  Your $M_2$ should not be zero, it should be $1 \pmod 4$ and $0 \pmod {9,7}$, which is $-63 \pmod {252}$  Your $M_3$ should be $36 \pmod {252}$  I don't know what your $y_1,y_2,y_3$ are.  That multiplication by $4$ is the source of your problem.  Then you have the solution is $x\equiv 7\cdot 28+0\cdot (-63)+2\cdot 36=268\equiv 16 \pmod {252}$
