Question 7

We can find the winning strategy by working backward from the target number.

Step 1 — Find the special winning numbers

The goal is to reach 99. The player who says 99 wins the game. To say 99, the number before it must be $X$. We must be able to add a number from 1 to 10 to $X$ to get 99. So, $X$ must be between $99 - 10 = \mathbf{89}$ and $99 - 1 = \mathbf{98}$. If the opponent says any number from 89 to 98, we can always add a number to reach 99. For example, if the opponent says 89, we add 10. If the opponent says 98, we add 1. All these additions are between 1 and 10. This means that if we can force the opponent to say a number from 89 to 98, we win. The number just before this range is 88. If we say 88, the opponent must say a number from 89 to 98. Then, we can add the correct number to reach 99. So, 88 is a special "winning number" for the player who says it.

We can find other special winning numbers by working backward from 88. The maximum number we can add is 10. So, the "magic step" is $10 + 1 = \mathbf{11}$. We subtract 11 from 88 to find the next special winning number. $88 - 11$ $= \mathbf{77}$ If we say 77, the opponent says a number between 78 and 87. Then, we can add the correct number to reach 88. For example, if the opponent says 78, we add 10. If the opponent says 87, we add 1. All these additions are between 1 and 10. So, 77 is also a special winning number. This pattern continues. The special winning numbers are multiples of 11. Let us list them: $\mathbf{11}, \mathbf{22}, \mathbf{33}, \mathbf{44}, \mathbf{55}, \mathbf{66}, \mathbf{77}, \mathbf{88}, \mathbf{99}$ The player who can always say these numbers can win the game.

$\boxed{\text{Winning numbers are } \mathbf{11, 22, ..., 99}}$

Step 2 — Which player can always win?

The first player says a number between 1 and 10. None of these numbers are multiples of 11. So, the first player cannot say a special winning number on their first turn. Let us say the first player says $N_1$. It is now the second player's turn. The current total is $N_1$. The second player wants to say the first special winning number, which is 11. The second player needs to add $k_2$ such that $N_1 + k_2 = 11$. So, the second player adds $k_2 = 11 - N_1$. Since $N_1$ is between 1 and 10, $k_2$ will also be between $11-10=\mathbf{1}$ and $11-1=\mathbf{10}$. This means the second player can always choose a valid number to add to reach 11. For example, if Player 1 says 5. Player 2 adds $11 - 5 = \mathbf{6}$. The total is now 11.

Now it is Player 1's turn. Player 1 must add a number $k_3$ between 1 and 10. The total will be $11+k_3$, which is between $11+1=\mathbf{12}$ and $11+10=\mathbf{21}$. None of these numbers are multiples of 11. So, Player 1 cannot say a special winning number.

Now it is Player 2's turn again. The current total is $T$. $T$ is not a multiple of 11. Player 2 wants to say the next special winning number, which is 22. Player 2 adds $k_4$ such that $T + k_4 = 22$. Player 2 adds $k_4 = 22 - T$. Since $T$ is between 12 and 21, $k_4$ will be between 1 and 10. This means the second player can always choose a valid number to add to reach 22.

This pattern continues throughout the game. The second player can always say 11, 22, 33, 44, 55, 66, 77, 88. When the second player says 88, it is the first player's turn. Player 1 adds a number $k_x$ between 1 and 10. The total will be $88+k_x$, which is between $88+1=\mathbf{89}$ and $88+10=\mathbf{98}$. None of these numbers are multiples of 11. Now it is the second player's turn. The current total is $T$. $T$ is not a multiple of 11. Player 2 wants to say the final special winning number, which is 99. Player 2 adds $k_y$ such that $T + k_y = 99$. Player 2 adds $k_y = 99 - T$. Since $T$ is between 89 and 98, $k_y$ will be between 1 and 10. The second player says 99 and wins the game.

$\boxed{\text{The second player can always win}}$

Answer

(i) The second player can always win. (ii) The winning player should always say numbers that are multiples of 11. These numbers are 11, 22, 33, 44, 55, 66, 77, 88, 99.