MOPSS, 21st February, 2026

By the claims below, it follows that it is possible to end with only one number remaining on the board if and only if \(n\) is a power of \(2\).

Claim. Let \(n\) be a positive integer. Assume that it is possible to end with only one number remaining on the board. Then \(n\) is a power of \(2\).

Proof of the claim. Let \(d\) be an odd positive divisor of \(n\). We will show that \(d\) must be \(1\). Since no two of the numbers \(1, 2, \dots, n\) differ by \(n\), the number \(n\) cannot be erased in any move. Therefore, \(n\) must be the last remaining number on the board.

Assume that \(d\) divides all the integers present on the board at some stage in the game. Denote these integers by \(a_1, a_2, \dots, a_k\). Denote by \(b\) the integer that was erased in the move prior to this stage.

Consider the case that \(b\) is the difference of \(b\) and \(a_i\) for some \(i\). Then \(b\) is equal to one of \(b - a_i, a_i - b\). Since \(a_i\) is positive, we obtain \(b = a_i - b\), which implies \(a_i = 2b\). Since \(d\) divides \(a_i\) and \(d\) is odd, it follows that \(d\) divides \(b\).

Now, consider the case that \(b\) is equal to \(a_i - a_j\) for some \(i\) and \(j\) with \(i \neq j\). Since \(d\) divides both \(a_i\) and \(a_j\), it follows that \(d\) divides \(b\).

This proves that if \(d\) divides all the integers present on the board at some stage, then \(d\) divides the integer that was erased in the move prior to this stage, that is, \(d\) divides all the integers present on the board prior to this stage as well. Since \(d\) divides \(n\), it follows that \(d\) divides all the integers present on the board at every stage prior to the last stage. In particular, \(d\) divides \(1\), which implies \(d = 1\).

This proves that \(n\) has no odd positive divisor other than \(1\). Therefore, \(n\) is a power of \(2\).

Claim. Let \(n\) be a positive integer. Assume that \(n\) is a power of \(2\). Then it is possible to end with only one number remaining on the board.

Proof of the claim. Write \(n = 2^k\) with \(k \geq 0\). Using induction, we will prove the claim. For the base case \(k = 0\), we have \(n = 1\). In this case, we already have only one number remaining on the board, so the claim holds. Also note that the claim holds for \(k = 1\) as well, since we can erase \(1\) and be left with \(2\).

Assume that the claim holds for some \(k \geq 1\), that is, for \(n = 2^k\). Consider \(n = 2^{k+1}\). We can divide the numbers \(1, 2, \dots, 2^{k+1}\) into two groups: \(1, 2, \dots, 2^k\) and \(2^k+1, 2^k+2, \dots, 2^{k+1}\). Note that \[ 2^k + i = 2^{k + 1} - (2^k - i) \] for any \(i\) with \(1 \leq i \leq 2^k - 1\). Therefore, we can erase the numbers \(2^k + 1, 2^k + 2, \dots, 2^{k+1} - 1\) by using the numbers \(1, 2, \dots, 2^k - 1\) and \(2^{k + 1}\). Thus, we can reduce the numbers on the board to \(1, 2, \dots, 2^k\) and \(2^{k+1}\). By the induction hypothesis, we can reduce \(1, 2, \dots, 2^k\) to \(2^k\). Finally, when we are left with \(2^k\) and \(2^{k+1}\) on the board, we can erase \(2^k\) and be left with \(2^{k+1}\) as the only number remaining on the board. By induction, the claim holds for all \(k \geq 0\).

Let us allow Bob to play along with Amber. As Amber fills a square, Bob fills the same square with a pair \((r, c)\) where \(r\) is the number of filled squares in that row before Amber fills it, and \(c\) is the number of filled squares in that column before Amber fills it. Note that the number Amber fills in that square is \(r + c\). After Amber fills all the squares, Bob will have filled all the squares with pairs of numbers. Let us denote the sum of all the numbers Amber filled in as \(S\). We have \[ S = \sum_{(r, c)} (r + c) = \sum_{(r, c)} r + \sum_{(r, c)} c. \] Now, let us look at the first sum \(\sum_{(r, c)} r\). For each row, the value of \(r\) across the squares in that row is \(0, 1, 2, \ldots, 9\) in some order. Therefore, the sum of \(r\) across the squares in that row is \(0 + 1 + 2 + \cdots + 9 = 45\). Since there are \(10\) rows, we have \[ \sum_{(r, c)} r = 10 \cdot 45 = 450. \] By a similar argument, we have \[ \sum_{(r, c)} c = 10 \cdot 45 = 450. \] Therefore, we have \[ S = \sum_{(r, c)} r + \sum_{(r, c)} c = 450 + 450 = 900. \] Thus, the only possible result of the addition is \(900\).

Let \(n > 1\) be a positive integer such that there is an assignment of the numbers \(1,2,\dots,n\) to the rocks such that no two frogs land on the same rock after the signal. Label the rocks as the first, second, …, \(n\)-th rock in the clockwise order. For each \(i \in \{1,2,\dots,n\}\), let \(a_i\) be the number written on the \(i\)-th rock. Then, after the signal, the frog on the \(i\)-th rock jumps to the \((i + a_i)\)-th rock (where the integer \(i + a_i\) is taken modulo \(n\)). Since no two frogs land on the same rock, the numbers \(i + a_i\) for \(i = 1,2,\dots,n\) are distinct modulo \(n\). In particular, the numbers \(i + a_i\) for \(i = 1,2,\dots,n\) are congruent to the numbers \(1,2,\dots,n\) in some order modulo \(n\). Therefore, we have \[ (1 + a_1) + (2 + a_2) + \cdots + (n + a_n) \equiv 1 + 2 + \cdots + n \pmod{n}, \] which implies that \(a_1 + a_2 + \cdots + a_n \equiv 0 \pmod{n}\). Since \(a_1 + a_2 + \cdots + a_n = 1 + 2 + \cdots + n = \frac{n(n+1)}{2}\), we have \(\frac{n(n+1)}{2} \equiv 0 \pmod{n}\). This implies that \(n+1\) is even, so \(n\) is odd. Conversely, let \(n > 1\) be an odd positive integer. Consider the assignment of the numbers \(1,2,\dots,n\) to the rocks in the order \(1,2,\dots,n\) in the clockwise direction. Then, after the signal, the frog on the \(i\)-th rock jumps to the \((i + i)\)-th rock, which is the \((2i)\)-th rock modulo \(n\). Since \(n\) is odd, the numbers \(2i\) for \(i = 1,2,\dots,n\) are distinct modulo \(n\). Therefore, no two frogs land on the same rock after the signal. Hence, the integers \(n > 1\) for which it is possible to assign the numbers \(1,2,\dots,n\) to the rocks in such a way that no two frogs land on the same rock after the signal are precisely the odd integers.

The first player, Daniel has a winning strategy. The strategy of Daniel is to remove a yellow stone during his turns. We claim that Daniel will win by following this strategy. More precisely, we will show that Daniel will win when only one yellow stone is left or earlier. Assume that after \(k\) turns in total, only one yellow stone is left. Assume that among these \(k\) turns, Daniel has removed \(x\) yellow stones, and Juan David has removed \(y\) yellow stones, \(b\) blue stones and \(r\) red stones. Note that after these \(k\) turns, there are \(30 - x - y = 1\) yellow stones, \(20 - b\) blue stones and \(10 - r\) red stones. Since Daniel is the first player, it follows that at any stage of the game, the number of turns of Juan David is either equal to or one less than the number of turns of Daniel. This implies that \(b + r \leq b + r + y \leq x \leq x + y = 29\). This shows that at least one of the inequalities \(b \leq 19\), \(r \leq 9\) holds. This gives that after \(k\) turns, at least one of the differences

\(\begin{align*} \text{the number of yellow stones} - \text{the number of blue stones} & = b - 19, \\ \text{the number of yellow stones} - \text{the number of red stones} & = r - 9 \end{align*} \\)

is less than or equal to zero. Observe that these two differences were both positive at the beginning of the game. Therefore, at some moment during the game, within these \(k\) turns, one of these two differences must have become zero. In other words, at some moment during the game, one of the red and blue piles had the same positive number of stones as the yellow pile. Noting that during these \(k\) turns, the number of yellow stones is always positive, we conclude that Daniel wins.

Let \(B\) denote the sum of the cubes of the numbers on the board at a moment, and \(N\) denote the sum of the numbers in Anya’s notes at the same moment. We will show that \(B - 3N\) is invariant under Anya’s operation. Let \(a\) and \(b\) be the numbers removed from the board. Note that \(B\) changes to \(B - a^3 - b^3 + (a + b)^3\) and \(N\) changes to \(N + ab(a + b)\). This shows that \(B - 3N\) changes to \[ B - a^3 - b^3 + (a + b)^3 - 3(N + ab(a + b)) = B - 3N. \] Since \(B - 3N\) is invariant, we have \[ 1^3 + 2^3 + \dots + 50^3 = (1 + 2 + \dots + 50)^3 - 3S, \] which gives

\(\begin{align*} S & = \frac{(1 + 2 + \dots + 50)^3 - (1^3 + 2^3 + \dots + 50^3)}{3} \\ & = \frac{(1 + 2 + \dots + 50)^3 - (1 + 2 + \dots + 50)^2}{3} \\ & = \frac{1}{3} \left( \left( \frac{50 \cdot 51}{2} \right)^2 \cdot \left( \frac{50 \cdot 51}{2} - 1 \right)\right) \\ & = \frac{1275^2 \cdot 1274}{3} . \end{align*} \\)

Let \(p\) denote the number of piles and \(s\) denote the total number of stones at some stage of the game. Note that after applying operation (i), we have \(p - 1\) piles and \(s + 2\) stones. Moreover, after applying operation (ii), we have \(p + 1\) piles and \(s - 2\) stones. This shows that \(2p + s\) remains invariant under both the operations. Let \(P\) denote the number of piles when the game has come to an end by performing the operations in a certain sequence. Note that there are at least \(P - 1\) piles containing only one stone. Let \(S\) denote the number of stones in the remaining pile. Observe that \(1 \leq S \leq 3\). The total number of stones is equal to \(S + (P - 1)\). This yields \[ 2P + S + P - 1 = 2 \times 10 + (1 + 2 + \dots + 10) , \] which implies \[ 3P + S = 2 \times 10 + (1 + 2 + \dots + 10) + 1 = 20 + 55 + 1 = 76. \] This implies that \(P = 25\) and \(S = 1\), so there are \(25\) piles with one stone in the end of the game.