[Algebra]
Problem: In an infinite array with two rows, the numbers in the top row are denoted $\ldots, A_{-2}, A_{-1}, A_{0}, A_{1}, A_{2}, \ldots$ and the numbers in the bottom row are denoted $\ldots, B_{-2}, B_{-1}, B_{0}, B_{1}, B_{2}, \ldots$ For each integer $k$, the entry $A_{k}$ is directly above the entry $B_{k}$ in the array, as shown:

| $\ldots$ | $A_{-2}$ | $A_{-1}$ | $A_{0}$ | $A_{1}$ | $A_{2}$ | $\ldots$ |
| :--- | :--- | :--- | :--- | :--- | :--- | :--- |
| $\ldots$ | $B_{-2}$ | $B_{-1}$ | $B_{0}$ | $B_{1}$ | $B_{2}$ | $\ldots$ |

For each integer $k, A_{k}$ is the average of the entry to its left, the entry to its right, and the entry below it; similarly, each entry $B_{k}$ is the average of the entry to its left, the entry to its right, and the entry above it.
In one such array, $A_{0}=A_{1}=A_{2}=0$ and $A_{3}=1$.

Determine the value of $A_{4}$.

Solution: We draw part of the array using the information that $A_{0}=A_{1}=A_{2}=0$ and $A_{3}=1$ :

$$
\begin{array}{l|l|l|l|l|l|l|lll|c|c|c|c|c|c}
\cdots & A_{0} & A_{1} & A_{2} & A_{3} & A_{4} & A_{5} & \cdots & \cdots & 0 & 0 & 0 & 1 & A_{4} & A_{5} & \cdots \\
\hline \cdots & B_{0} & B_{1} & B_{2} & B_{3} & B_{4} & B_{5} & \cdots & \cdots & B_{0} & B_{1} & B_{2} & B_{3} & B_{4} & B_{5} & \cdots
\end{array}
$$

Since $A_{1}$ is the average of $A_{0}, B_{1}$ and $A_{2}$, then $A_{1}=\frac{A_{0}+B_{1}+A_{2}}{3}$ or $3 A_{1}=A_{0}+B_{1}+A_{2}$. Thus, $3(0)=0+B_{1}+0$ and so $B_{1}=0$.

Since $A_{2}$ is the average of $A_{1}, B_{2}$ and $A_{3}$, then $3 A_{2}=A_{1}+B_{2}+A_{3}$ and so $3(0)=0+B_{2}+1$ which gives $B_{2}=-1$.

Since $B_{2}$ is the average of $B_{1}, A_{2}$ and $B_{3}$, then $3 B_{2}=B_{1}+A_{2}+B_{3}$ and so $3(-1)=0+0+B_{3}$ which gives $B_{3}=-3$.

So far, this gives

$$
\begin{array}{l|c|c|c|c|c|c|l}
\cdots & 0 & 0 & 0 & 1 & A_{4} & A_{5} & \cdots \\
\hline \cdots & B_{0} & 0 & -1 & -3 & B_{4} & B_{5} & \cdots
\end{array}
$$

Since $A_{3}$ is the average of $A_{2}, B_{3}$ and $A_{4}$, then $3 A_{3}=A_{2}+B_{3}+A_{4}$ and so $3(1)=$ $0+(-3)+A_{4}$ which gives $A_{4}=6$.

Final answer: 6