<s>[INST] 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
$$
\begin{split}
M &= 88 = 8 \times 11 \\
x_1 &= 123^{456}\equiv 2^{456} \equiv 2^{6} \equiv 64 \equiv 9 \pmod{11} \\
y_1 &= 9^{-1} \equiv 9^9 \equiv (-2)^9 \equiv -512 \equiv -6 \equiv 5 \pmod{11}\\
x_2 &= 123^{456} \equiv 123^0 \equiv 1 \pmod{8}\\
y_2 &= 1^{-1} \equiv 1 \pmod{8} \\
123^{456}
 &\equiv \sum_{i=1}^2 x_i\times\frac{M}{m_i} \times y_i
  \equiv 9\times\frac{88}{11}\times5 + 1\times\frac{88}{8} \times1 \equiv 371 
  \equiv 19 \pmod{88}
\end{split}
$$
 [/INST] Your calculations look correct except for the last line which I don't understand.
One you get $x_1$ and $x_2$, you could simply write
$x=123^{456}=9+11k$ (from $x_1$)
so reducing mod $8$ yields
$x \equiv 1+3k \pmod{8} \equiv 1 \pmod{8}$ (from $x_2$)
therefore
$3k\equiv0 \pmod{8}$ and since $\gcd(3,8)=1$, $3$ is invertible mod $8$, so $k \equiv 0 \pmod{8}$
hence $x=9+11(0+8k')=9+88k' \equiv 9 \pmod{88}$
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