Let $a$ be a natural number. What are the possible values of $a^{5}$ modulo 11?
Let $x=a^{5}$, and let's consider the values taken by $x$.

If $a$ is divisible by $11, x \equiv a^{5} \equiv 0(\bmod 11)$. Now suppose that $a$ and 11 are coprime. Then

$$
x^{2} \equiv a^{10} \equiv 1 \quad(\bmod 11)
$$

Therefore, $x$ is a solution to the equation

$$
x^{2}-1 \equiv 0 \quad(\bmod 11)
$$

We can factorize $x^{2}-1$ as $(x-1)(x+1)$, so

$$
11 \mid(x-1)(x+1)
$$

Since 11 is a prime number, 11 divides $x-1$ or $x+1$. Thus $x \equiv a^{5} \equiv \pm 1(\bmod 11)$. Finally, the values taken by $a^{5}$ modulo 11 are $-1,0,1$, and it is easy to verify that they are all achieved.