1. Verify that the number $7^{20}-1$ is divisible by $10^{3}$ and determine the last three digits of the number $7^{2015}$.
## Solution.

For $n \in \mathbb{N}$, we denote by $\mathcal{M} n$ a natural multiple of the number $n$. We have

$$
\begin{aligned}
7^{5} & =16807=\mathcal{M} 1000+807 \\
7^{10} & =\left(7^{5}\right)^{2}=\mathcal{M} 1000+807^{2}=\mathcal{M} 1000+651249=\mathcal{M} 1000+249 \\
7^{20} & =\left(7^{10}\right)^{2}=\mathcal{M} 1000+249^{2}=\mathcal{M} 1000+62001=\mathcal{M} 1000+1
\end{aligned}
$$

Thus, $7^{20}-1$ is divisible by $10^{3}$. (4p)

Then

$$
\begin{aligned}
7^{2015} & =7^{2000} \cdot 7^{15}=\left(7^{20}\right)^{100} \cdot\left(7^{5}\right)^{3}=(\mathcal{M} 1000+1)^{100} \cdot(\mathcal{M} 1000+807)^{3} \\
& =(\mathcal{M} 1000+1) \cdot(\mathcal{M} 1000+525557943)=(\mathcal{M} 1000+1) \cdot(\mathcal{M} 1000+943)=\mathcal{M} 1000+943
\end{aligned}
$$

Therefore, the last three digits of the number $7^{2015}$ are 943. (3p)