3. In a square table with 2015 rows and columns, positive numbers are arranged. The product of the numbers in each row and each column is 2, and the product of the numbers in any $3 \times 3$ square is 1. What number is in the center of the table?
# Solution

Consider the first 3 rows of the table. From the additional condition, it follows that if these rows are covered "butt-to-butt" with squares of size $3 \times 3$, moving simultaneously from left and right towards each other, then the 336-th square from the left and the 336-th square from the right will overlap by one column (since $(336 \cdot 3) \cdot 2 = 2015 + 1$). Denote the product of the numbers in this column (there are 3 of them) by $M$.

Then the product of all numbers in the first 3 rows of the table is, on the one hand, $2^{3}$, and on the other hand, $1 / M$.

Thus, $\quad M = \frac{1}{2^{3}}$.

Now consider the middle (1008-th) column of the table. It is similarly divided into $336 \cdot 2$ blocks of 3 elements, which overlap at the central element of the table (if moving from top and bottom towards each other). Denote this element by $C$.

The product of all numbers in this column is 2. Therefore, $2 = M^{2 \cdot 336} / C$.

From which $C = \frac{1}{2} \cdot \frac{1}{2^{2 \cdot 336 \cdot 3}} = 2^{-2017}$.

Answer. The number in the center of the table is $2^{-2017}$.