Exercise 1.3.3 (Divisibility criterion by $2,4,8$)

Proof. Let $n=\overline{a_m{a}_{m-1}\cdots a_0}={\sum <!--\; FUNCTION\; APPLICATION\; -->}_{i=0}^m{a}_{i}10^i$ a positive integer, written into base $10$.

Since $10\equiv 0(\mathit{mod}2)$, $n\equiv a_0(\mathit{mod}2)$, so $2\mid n-a_0$. If $2\mid a_0$, then $2\mid n$, and if $2\mid n$ then $2\mid a_0$: $n$ is divisible by $2$ if and only if its unit digit is divisible by $2$.

Since $100=10^2\equiv 0(\mathit{mod}4)$, $n\equiv a_0+10a_1(\mathit{mod}4)$, so $4\mid n-(a_0+10a_1)$. So $n$ is divisible by $4$ if and only if the integer $\overline{a_1{a}_0}=a_0+10a_1$ formed by its tens digits and its unit digit is divisible by $4$

Since $1000=10^3\equiv 0(\mathit{mod}8)$, $n\equiv a_0+10a_1+100a_1(\mathit{mod}4)$, so $8\mid n-(a_0+10a_1+100a_2)$. So $n$ is divisible by $8$ if and only if the integer $\overline{a_2{a}_1{a}_0}=a_0+10a_1+100a_2$ formed by its last three digits is divisible by $8$. □