55rd Ukrainian National Mathematical Olympiad - Third Round

a) Let us show that the nearest sequence of $7$ colored years is $2103,\dots ,2109$. First, we prove that in this century no sequence of more than six colored years can happen any longer. We see that digits $0$ and $2$ cannot represent units or tens. Therefore, a chain is broken at each year ending in $0$ or $2$, which gives us the only way to form a chain of $7$ years: $$20\ast 3,20\ast 4,\dots ,20\ast 9.$$ Here, $\ast$ can only be equal to $1$, but this is the current chain. In the next century, after $2100$, the first chain is easy to find: $2103,\dots ,2109$.

b) Years we deal with are written with at least four digits. A colored chain cannot contain numbers ending in $99$, hence the number of hundreds doesn't change throughout the chain. This implies that there are only $8$ possible last digits. Assume some colored chain has $8$ numbers. We have two cases. 1. All years have the same number of tens. Then there are only $7$ options for the last digit. Contradiction. 2. The number of tens changes within the chain. In this case the tens digit cannot be equal to $9$. Assume this digit changes from $x$ to $x+1$. The chain has $8$ numbers, thus, it should have two numbers of the type $$\overline{ab}x+x+1,\overline{ab}x+1x.$$ For example, they might be $2145$ and $2154$. But then the chain has at least $10$ years (for example, $2145$ to $2154$). Again, we have a contradiction.