6. If $(2 x+4)^{2 n}=a_{0}+a_{1} x+a_{2} x^{2}+\cdots+a_{2 n} x^{2 n}\left(n \in \mathbf{N}^{*}\right)$, then the remainder when $a_{2}+a_{4}+\cdots+a_{2 n}$ is divided by 3 is $\qquad$
Answer 1.
Analysis Let $x=0$, we get $a_{0}=4^{2 n}$; by letting $x=1$ and $x=-1$ respectively, and adding the two resulting equations, we get
$$
a_{0}+a_{2}+a_{4}+\cdots+a_{2 n}=\frac{1}{2}\left(6^{2 n}+2^{2 n}\right)
$$

Therefore,
$$
a_{2}+a_{4}+\cdots+a_{2 n}=\frac{1}{2}\left(6^{2 n}+2^{2 n}\right)-4^{2 n}=2^{2 n-1}\left(3^{2 n}+1\right)-4^{2 n}=(-1)^{2 n-1} \times 1-1^{2 n}=-2=1 \quad(\bmod 3)
$$