Divisibility rules

Divisibility by $2$ and $5$. Since $10\equiv0\pmod2$ and $\pmod5$, only the last digit matters: $n\equiv a_0\pmod2$ and $\pmod5$. So $2\mid n$ iff the last digit is even; $5\mid n$ iff the last digit is $0$ or $5$.

Divisibility by $3$ and $9$. Since $10\equiv1\pmod3$ and $\pmod9$, $n\equiv\sum a_k\pmod3$ (and mod $9$). So $3\mid n$ (resp. $9\mid n$) iff the digit sum is divisible by $3$ (resp. $9$).

Divisibility by $11$. Since $10\equiv-1\pmod{11}$, $n\equiv\sum(-1)^ka_k\pmod{11}$. So $11\mid n$ iff the alternating digit sum is divisible by $11$.

Conclusion. These rules follow directly from $10\equiv0,1,-1$ modulo the respective divisors.