## Task B-3.3.

Determine how many pairs of natural numbers $(x, y)$ satisfy the equation

$$
\log _{2}\left(\log _{2^{x}}\left(\log _{2^{y}} 2^{10^{2020}}\right)\right)=0
$$

What is the largest product $x y$ among all such pairs?
## Solution.

The given equation is equivalent to $\log _{2^{x}}\left(\log _{2^{y}} 2^{10^{2020}}\right)=1$, or $\log _{2^{x}}\left(\log \frac{10^{2020}}{y}\right)=1$. It follows that $\frac{10^{2020}}{y}=2^{x}$, so

$$
y=\frac{10^{2020}}{2^{x}}=\frac{2^{2020} \cdot 5^{2020}}{2^{x}}=2^{2020-x} \cdot 5^{2020}
$$

Since $x$ and $y$ are natural numbers, $2020-x>0$ must hold, so $x<2020$, or $x \in$ $\{1,2,3, \ldots, 2020\}$. Therefore, there are 2020 desired ordered pairs

$$
(x, y)=\left(x, 2^{2020-x} \cdot 5^{2020}\right)
$$

The product $x y=x \cdot 2^{2020-x} \cdot 5^{2020}$ is maximized when the product $x \cdot 2^{-x}$ is maximized.

Consider the difference between the products $x \cdot 2^{-x}$ and $(x+1) \cdot 2^{-(x+1)}$.

$$
x \cdot 2^{-x}-(x+1) \cdot 2^{-(x+1)}=2^{-x-1}(2 x-x-1)=2^{-x-1}(x-1)
$$

Clearly, this difference is positive for all $x \geqslant 1$, which means that the value of the product $x \cdot 2^{-x}$ decreases as the number $x$ increases, so its value is greatest for $x=1$.

Therefore, the maximum value of the product is $2^{2019} \cdot 5^{2020}=\frac{10^{2020}}{2}$.