Japan Mathematical Olympiad

2883

Let $k$ be a positive integer. Consider arranging black and white stones in a single row, with a total of $5^k$ stones. Define a bad arrangement as one where, after $k$ repetitions of an operation, the last remaining stone is white.

Define $f(k)$ as the minimum non-negative integer $m$ such that a bad arrangement can be formed with $5^k-m$ black stones and $m$ white stones. Obviously, $f(1)=3$. If $m\ge f(k)$, then there exists a bad arrangement with $5^k-m$ black stones and $m$ white stones.

Now, if $n\ge 5^5-f(5)+1$, then there are at most $f(5)-1$ white stones, and since no bad arrangement exists, the last remaining stone must be a black stone. Conversely, if $n\le 5^5-f(5)$, then there are $f(5)$ or more white stones, and there exists an arrangement where the last remaining stone is a white stone. Therefore, the desired value is $5^5-f(5)+1$. Thus, we will consider expressing $f(k+1)$ in terms of $f(k)$ for $k\ge 1$.

Consider a bad arrangement $W_k$ with $5^k-f(k)$ black stones and $f(k)$ white stones. Now, arrange $2\cdot 5^k$ black stones in the beginning, followed by three instances of $W_k$, forming a row of $5^{k+1}$ stones. The number of white stones in this row is $3f(k)$. After $k$ repetitions of the operation, the final arrangement consists of black, black, white, white, white stones, ensuring that the last remaining stone is white. Therefore, $f(k+1)\le 3f(k)$.

Conversely, consider a row of $5^{k+1}$ stones where there are at most $3f(k)-1$ white stones. Divide this row into sets of $5^k$ stones each. Since there are fewer than $3f(k)$ white stones, there can be at most 2 sets with $f(k)$ or more white stones. From the minimality of $f(k)$, after $k$ repetitions of the operation, there can be at most 2 white stones among the last 5 remaining stones. Thus, the last remaining stone is black, implying $f(k+1)>3f(k)-1$.

Hence, we have $f(k+1)=3f(k)$. Since $f(5)=3^4\cdot f(1)=3^5$, the desired value is $5^5-f(5)+1=2883$.