[b]p1.[/b] One throws randomly $120$ squares of the size $1\times 1$ in a $20\times 25$ rectangle. Prove that one can still place in the rectangle a circle of the diameter equal to $1$ in such a way that it does not have common points with any of the squares.


[b]p2.[/b] How many digits has the number $2^{70}$ (product of $70$ factors of $2$)?


[b]p3.[/b] Prove that the equation $x^2 - 2y^2 = 1$ has infinitely many integer solutions (i.e. there are infinitely many integers $x$ and $y$ which satisfy this equation).


[b]p4.[/b] Find all $x, y, z$ which satisfy the following system of equations:
$$\frac{xy}{x + y}=\frac83$$
$$\frac{zy}{z + y}=\frac{12}{5}$$
$$\frac{zx}{z + x}=\frac{24}{7}$$


PS. You should use hide for answers.
To determine the number of digits in \(2^{70}\), we can use the formula for the number of digits of a number \(n\), which is given by:

\[
\text{Number of digits of } n = \lfloor \log_{10} n \rfloor + 1
\]

In this case, \(n = 2^{70}\). Therefore, we need to find \(\lfloor \log_{10} (2^{70}) \rfloor + 1\).

1. Calculate \(\log_{10} (2^{70})\):
   \[
   \log_{10} (2^{70}) = 70 \log_{10} 2
   \]

2. Use the known value \(\log_{10} 2 \approx 0.3010\):
   \[
   70 \log_{10} 2 \approx 70 \times 0.3010 = 21.07
   \]

3. Apply the floor function and add 1:
   \[
   \lfloor 21.07 \rfloor + 1 = 21 + 1 = 22
   \]

Thus, the number \(2^{70}\) has 22 digits.

The final answer is \(\boxed{22}\).