33rd Hellenic Mathematical Olympiad

The sum ${\alpha }_1+{\alpha }_2+{\alpha }_3+{\alpha }_4+{\alpha }_5+{\alpha }_6$ is even, if and only if, the number of $1$'s is even, that is $0$, $2$, $4$, $6$.

In the case of zero $1$'s, the possible selections are $2^6$, because for each ${\alpha }_i$ we have $2$ selections, ($0$ or $2$).

When we have two $1$'s, then they can be selected in $\left(\genfrac{}{}{0ex}{}{6}{2}\right)$ ways and the rest four places can be completed by $2^4$ ways. It means that we have $2^4\cdot \left(\genfrac{}{}{0ex}{}{6}{2}\right)$ possible $6$-tuples.

Similarly, in the case we have four $1$'s we conclude that we have $2^2\cdot \left(\genfrac{}{}{0ex}{}{6}{4}\right)$ possible $6$-tuples.

When we have six $1$'s then obviously we have only one possible $6$-tuple.

Therefore the possible $6$-tuples are totally: $$2^6+2^4\cdot \left(\genfrac{}{}{0ex}{}{6}{2}\right)+2^2\cdot \left(\genfrac{}{}{0ex}{}{6}{4}\right)+1=64+16\cdot 15+4\cdot 15+1=365.$$