343. Let's write down the consecutive powers of the number 2:

$2,4,8,16,32,64,128,256,512,1024,2048,4096, \ldots$

It is easy to notice that the last digit in this sequence of numbers repeats periodically with a period of 4:

$$
2,4,8,6,2,4,8,6,2,4,8,6, \ldots
$$

Prove that the last 10 digits in this sequence of numbers, starting from some number, will also repeat periodically. Find the length of the period and the number in the sequence, starting from which the periodicity is observed.
343. By Euler's theorem (see problem 341), the number $2^{5^{50}-5^{0}}-1=$ $=24 \cdot 5^{9}-1=2^{7812500}-1$ is divisible by $5^{10}$; therefore, for $n \geqslant$ $\geqslant 10$ the difference $2^{7812500+n}-2^{n}=2^{n}\left(2^{7812500}-1\right)$ is divisible by $10^{10}$, i.e., the last 10 digits of the numbers $2^{7812500+n}$ and $2^{n}$ coincide. This means that the last 10 digits of the numbers in the sequence $2^{1}, 2^{2}, 2^{3}, \ldots, 2^{n}, \ldots$ repeat every 7812500 numbers, and this periodicity begins with the tenth number in this sequence - with the number $2^{10}$.

The fact that the period is not less than 7812500 follows from the result of problem 342.

Note. Similarly, it can be proved that the last $n$ digits of the numbers in the considered sequence repeat every $4 \cdot 5^{n-1}$ numbers, starting from the $n$-th number in this sequence (for example, the last two digits repeat, starting from the second number, every 20 numbers).