5. How many even six-digit numbers exist, in the notation of which identical digits do not stand next to each other
Solution. Let $N_{k}(k>1)$ denote the number of even $k$-digit numbers in which identical digits do not stand next to each other. Similarly, define $N_{1}$, but exclude the number 0, so $N_{1}=4$. We can directly calculate $N_{2}=41$. Let $k>2$. Notice that from $k$-digit numbers satisfying the condition, we can form suitable $(k+1)$-digit numbers by adding one of the 8 permissible digits (not zero and not the one in the highest place) to the left. In addition, $(k+1)$-digit numbers can be formed from $(k-1)$-digit numbers by adding one of the 9 non-zero digits and zero to the left. There are no other options. From this, we obtain the recurrence relation: $N_{k+1}=8 \cdot N_{k}+9 \cdot N_{k-1}$. Sequentially calculating the values using this formula, we find the answer: $N_{6}=265721$.

Remark. One can also construct a proof based on the idea that among numbers in which identical digits do not stand next to each other, there are roughly equal numbers of even and odd numbers (for example, the correspondence is built by replacing digits according to the rule $d \rightarrow 9-d$). Specifically, it can be proven that when the number of digits is odd, there is one more even number than odd, and vice versa when the number of digits is even. However, this fact also requires a careful proof.