Find the number ot 6-tuples $(x_1, x_2,...,x_6)$, where $x_i=0,1 or 2$ and $x_1+x_2+...+x_6$ is even
To find the number of 6-tuples \((x_1, x_2, \ldots, x_6)\) where \(x_i = 0, 1, \text{ or } 2\) and the sum \(x_1 + x_2 + \cdots + x_6\) is even, we need to consider the parity of the sum. Specifically, the sum will be even if and only if there is an even number of 1s in the tuple.

1. **Case 1: 0 ones**
   - If there are 0 ones, all \(x_i\) must be either 0 or 2.
   - Each \(x_i\) has 2 choices (0 or 2).
   - Therefore, the number of such tuples is \(2^6\).

2. **Case 2: 2 ones**
   - If there are 2 ones, we need to choose 2 positions out of 6 for the ones.
   - The number of ways to choose 2 positions out of 6 is \(\binom{6}{2}\).
   - The remaining 4 positions can be either 0 or 2, giving \(2^4\) choices.
   - Therefore, the number of such tuples is \(\binom{6}{2} \cdot 2^4\).

3. **Case 3: 4 ones**
   - If there are 4 ones, we need to choose 4 positions out of 6 for the ones.
   - The number of ways to choose 4 positions out of 6 is \(\binom{6}{4}\).
   - The remaining 2 positions can be either 0 or 2, giving \(2^2\) choices.
   - Therefore, the number of such tuples is \(\binom{6}{4} \cdot 2^2\).

4. **Case 4: 6 ones**
   - If there are 6 ones, all \(x_i\) must be 1.
   - There is exactly 1 such tuple.

Summing up all these cases, we get the total number of 6-tuples where the sum is even:
\[
2^6 + \binom{6}{2} \cdot 2^4 + \binom{6}{4} \cdot 2^2 + 1
\]

Now, let's calculate each term:
\[
2^6 = 64
\]
\[
\binom{6}{2} = \frac{6!}{2!(6-2)!} = 15 \quad \text{and} \quad 15 \cdot 2^4 = 15 \cdot 16 = 240
\]
\[
\binom{6}{4} = \frac{6!}{4!(6-4)!} = 15 \quad \text{and} \quad 15 \cdot 2^2 = 15 \cdot 4 = 60
\]
\[
1
\]

Adding these together:
\[
64 + 240 + 60 + 1 = 365
\]

The final answer is \(\boxed{365}\)