. Determine the units digit of $2^{49}$.

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. On a $: 2^{1} \equiv 2(\bmod 10), 2^{2} \equiv 4(\bmod 10), 2^{3} \equiv 8(\bmod 10)$, $2^{4} \equiv 6(\bmod 10)$ and $2^{5} \equiv 2(\bmod 10)$.

We can then prove that $\forall k \in \mathbb{N}^{*}, 2^{4 k} \equiv 6(\bmod 10)$. Let's reason by induction on $k \in \mathbb{N}^{*}$. For $k=1$, we have: $2^{4 k}=2^{4} \equiv 6(\bmod 10)$. Suppose that: $2^{4 k} \equiv 6(\bmod 10)$ for $k \in \mathbb{N}^{*}, 2^{4(k+1)}=2^{4 k} \times 2^{4} \equiv 6(\bmod 10)$, which completes the induction. It follows that $\forall k \in \mathbb{N}, 2^{4 k+1} \equiv 2(\bmod 10)$ (indeed, for $k=0,2^{1} \equiv 2(\bmod 10)$ and for $k \in \mathbb{N}^{*}, 2^{4 k+1}=2^{4 k} \times 2$ and for $k \in \mathbb{N}^{*}$, $\left.2^{4 k} \equiv 6(\bmod 10)\right), 2^{4 k+2} \equiv 4(\bmod 10)$ and $2^{4 k+3} \equiv 8(\bmod 10)$. Perform the Euclidean division of 49 by 4. We have: $49=4 \times 12+1=4 k+1$ with $k=12$. Thus, $2^{49} \equiv 2(\bmod 10)$. Therefore, the units digit of $2^{49}$ is 2.

We could have reasoned more simply by writing that $2^{49}=2^{10} \times 2^{10} \times 2^{10} \times$ $2^{10} \times 2^{9}$. Since $2^{10}=\left(2^{5}\right)^{2}$ and $2^{5} \equiv 2(\bmod 10)$, we have: $2^{10} \equiv 4(\bmod 10)$. Furthermore, $2^{4} \equiv 6(\bmod 10)$, so $2^{9} \equiv 2(\bmod 10)\left(\right.$ by writing that $\left.2^{9}=\left(2^{4}\right)^{2} \times 2\right)$. Thus,
$2^{10} \times 2^{10} \equiv 6(\bmod 10)$ and $2^{10} \times 2^{10} \times 2^{10} \times 2^{10} \times 2^{9} \equiv 2(\bmod 10)$.