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Let $f(n)$ denote the number of $n-$ tuples $(x_1, . . . ,x_n)$ such that $0\le x_1, ... , x_n ≤ 7$ and $\displaystyle 5\mid\sum_{i=1}^n 3^{x_i}$ .

To compute $f(n+1)$ from $f(n)$ , we note that given any $n-$ tuple $(x_1, ... , x_n)$ such that $0\le x_1, ... , x_n ≤ 7$ and $\displaystyle 5\nmid\sum_{i=1}^n 3^{x_i}$ , there are exactly two possible values for $x_{n+1}$ such that $0 \le x_{n+1}\le 7$ and $\displaystyle 5\mid\sum_{i=1}^{n+1} 3^{x_i}$ , because $3^n \equiv 1, 3, 4, 2, 1, 3, 4, 2 \pmod{5}$ for $n = 0, 1, 2, 3, 4, 5, 6, 7$ respectively.

Also, given any valid $(n + 1)-$ tuple $(x_1, . . . , x_{n+1})$ , we can remove $x_{n+1}$ to get an $n-$ tuple $(x_1, . . . , x_n)$ such that such that $0\le x_1, ... , x_n ≤ 7$ and $\displaystyle 5\nmid\sum_{i=1}^n 3^{x_i}$ ., so these are in bijection.

There are a total of $8^n$ $n-$ tuples, $f(n)$ of which satisfy $\displaystyle 5\mid\sum_{i=1}^n 3^{x_i}$ , so there are $8^n - f(n)$ for which $\displaystyle 5\nmid\sum_{i=1}^n 3^{x_i}$ .

Therefore, $f(n + 1) = 2(8^n - f(n))$ .

We now have $f(1) = 0, f(2) = 2(8^1 - 0) = 16, f(3) = 2(8^2 - 16) = 96,\\ f(4) = 2(8^3 - 96) = 832, f(5) = 2(8^4 - 832) = 6528,\\ f(6)=2(8^5-6528)=52480, f(7)=2(8^6-52480)=\boxed{419328}$ .