B4. In the square of numbers below, positive numbers are listed. The product of the numbers in each row, each column, and each of the two diagonals is always the same.

What number is in the position of $H$?

| $\frac{1}{2}$ | 32 | $A$ | $B$ |
| :---: | :---: | :---: | :---: |
| $C$ | 2 | 8 | 2 |
| 4 | 1 | $D$ | $E$ |
| $F$ | $G$ | $H$ | 16 |
B4.

The product of the eight numbers in the second and fourth rows is equal to the product of the eight numbers in the first and second columns. That is,

$$
C \cdot 2 \cdot 8 \cdot 2 \cdot F \cdot G \cdot H \cdot 16=\frac{1}{2} \cdot C \cdot 4 \cdot F \cdot 32 \cdot 2 \cdot 1 \cdot G .
$$

| $\frac{1}{2}$ | 32 | 8 | 1 |
| :---: | :---: | :---: | :---: |
| 4 | 2 | 8 | 2 |
| 4 | 1 | 8 | 4 |
| 16 | 2 | $\frac{1}{4}$ | 16 |

Since $C, F$, and $G$ are not zero, we can divide both sides by $C, F$, and $G$. We then find $512 \cdot H=128$, and thus $H=\frac{1}{4}$. The solution shown above is a possible solution with $H=\frac{1}{4}$.