Task B-4.2. Let $n$ be the number obtained by writing 2013 zeros between every two digits of the number 14641. Determine all solutions of the equation $x^{4}=n$ in the set $\mathbb{C}$.
## Solution.

If between every two digits of the number 14641 we write 2013 zeros, we get a number of the form $1 \underbrace{0 \ldots 0}_{2013} 4 \underbrace{0 \ldots 0}_{2013} 6 \underbrace{0 \ldots 0}_{2013} 4 \underbrace{0 \ldots 0}_{2013} 1$, and we can express it using powers with base 10. Then,

$$
\begin{aligned}
n & =1 \underbrace{0 \ldots 0}_{2013} 4 \underbrace{0 \ldots 0}_{2013} 6 \underbrace{0 \ldots 0}_{2013} 4 \underbrace{0 \ldots 0}_{2013} 1 \\
& =10^{4 \cdot 2013+4}+4 \cdot 10^{3 \cdot 2013+3}+6 \cdot 10^{2 \cdot 2013+2}+4 \cdot 10^{2013+1}+1= \\
& =\left(10^{2013+1}\right)^{4}+4 \cdot\left(10^{2013+1}\right)^{3}+6 \cdot\left(10^{2013+1}\right)^{2}+4 \cdot 10^{2013+1}+1= \\
& =\binom{4}{0}\left(10^{2014}\right)^{4}+\binom{4}{1}\left(10^{2014}\right)^{3} \cdot 1+\binom{4}{2}\left(10^{2014}\right)^{2} \cdot 1^{2}+\binom{4}{3} 10^{2014} \cdot 1^{3}+\binom{4}{4} \cdot 1^{4}
\end{aligned}
$$

Then the solutions to the equation $x^{4}=n$, or $x^{4}=\left(10^{2014}+1\right)^{4}$, are the numbers

$$
x_{1,2}= \pm\left(10^{2014}+1\right), \quad x_{3,4}= \pm\left(10^{2014}+1\right) \cdot i
$$