Number Play - Games and Winning Strategies

Find the values of $A$ and $B$ in the following addition: $\begin{array}{r} 31A \\ + 1A3 \\ \hline 501 \end{array}$

From the units column: $A + 3 = 11$ (since it ends in $1$ and must be greater than $3$). This gives $A = 8$. Carry over $1$ to the tens column. Tens column: $1 (\text{carry}) + 1 + A = 1 + 1 + 8 = 10$. This gives $0$ in the result and a carry of $1$. Hundreds column: $1 (\text{carry}) + 3 + 1 = 5$, which matches. Thus, $A = 8$.

We use the rules of vertical addition. Since $A + 3$ results in a units digit of $1$, $A$ must be $8$. We then verify the carry-over for the subsequent columns.

If $21y5$ is a multiple of $9$, where $y$ is a digit, what is the value of $y$?

For a number to be divisible by $9$, the sum of its digits must be a multiple of $9$. Sum of digits $= 2 + 1 + y + 5 = 8 + y$. For $8 + y$ to be a multiple of $9$, $8 + y$ could be $9, 18, \dots$. Since $y$ is a single digit, $8 + y = 9$ is the only possibility. Therefore, $y = 9 - 8 = 1$.

We applied the divisibility rule for $9$: the sum of digits $2+1+y+5 = 8+y$ must be divisible by $9$. The nearest multiple of $9$ is $9$ itself.

Two players are playing a game where they add $1, 2, 3,$ or $4$ to a running total. The player who reaches exactly $30$ wins. If the current total is $20$, what should the next player add to ensure a win?

The winning positions are calculated by subtracting $(n+1)$ from the target, where $n=4$. The target is $30$. The steps are $30 - 5 = 25$, $25 - 5 = 20$, $20 - 5 = 15$. Since the current total is $20$, the current player is in a 'losing position' if the opponent plays correctly. However, if the player wants to reach the next target of $25$, they should add $25 - 20 = 5$. But the max they can add is $4$. The strategy is to always aim for the sequence $30, 25, 20, 15, 10, 5$. If the total is $20$, the next player cannot reach $25$ in one move.

In games with steps $1$ to $n$, the winning strategy involves controlling the remainder of the sum modulo $(n+1)$.