Finding remainder of $123^{456}$ divided by 88 using Chinese Remainder Theorem I tried using Chinese remainder theorem but I kept getting 19 instead of 9.
Here are my steps
$$
\begin{split}
M &= 88 = 8 \times 11 \\
x_1 &= 123^{456}\equiv 2^{456} \equiv 2^{6} \equiv 64 \equiv 9 \pmod{11} \\
y_1 &= 9^{-1} \equiv 9^9 \equiv (-2)^9 \equiv -512 \equiv -6 \equiv 5 \pmod{11}\\
x_2 &= 123^{456} \equiv 123^0 \equiv 1 \pmod{8}\\
y_2 &= 1^{-1} \equiv 1 \pmod{8} \\
123^{456}
 &\equiv \sum_{i=1}^2 x_i\times\frac{M}{m_i} \times y_i
  \equiv 9\times\frac{88}{11}\times5 + 1\times\frac{88}{8} \times1 \equiv 371 
  \equiv 19 \pmod{88}
\end{split}
$$

You used an incorrect CRT formula. It should be: for coprime $\,m,n,\,$ and $\,c^{-1}_{\ n}:= c^{-1}\bmod n$
$\qquad\begin{align} &x\equiv a\!\!\pmod{\!m}\\ &x\equiv b\!\!\pmod{\!n}\end{align}\iff x\,\equiv\, a\,n(n^{-1}_{\ m}) + b\,m(m^{-1}_{\ n})\ \ \pmod{\!mn}$
To help remember the formula note that it is easy to see it is correct since
$\qquad\qquad\qquad\quad\ \ \bmod \color{#c00} m\!:\ \  x\,\equiv\, a\, \underbrace{n (n^{-1})}_{\large \equiv\ 1\ }\, +\, \underbrace{b\, \color{#c00}m\, (\cdots)}_{\large \color{#c00}{\equiv\ 0}}\, \equiv\, a$
i.e. $\bmod (m,n)\!:\,\ n\, n^{-1}_{\ m}\equiv (1,0),\,\ m\,m^{-1}_{\ n}\equiv (0,1)\,$ so the formula is simply
$$ x\equiv (a,b) \equiv  a (1,0) + b (0,1)\qquad\qquad$$
See this answer for further discussion of this viewpoint (including an example with $3$ moduli).
