Alternating digit-sum test for divisibility by 11

Step 1. Show by induction that $10^i \equiv (-1)^i \pmod{11}$ for every $i\ge0$. Base case $i=0$: $10^0=1=(-1)^0$. Inductive step: if $10^i\equiv(-1)^i\pmod{11}$, then $10^{i+1}=10\cdot10^i\equiv(-1)\cdot(-1)^i=(-1)^{i+1}\pmod{11}$ (using $10 \equiv -1 \pmod{11}$). So $10^i\equiv(-1)^i\pmod{11}$ for all $i$.

Step 2. Substitute into the place-value expansion: $n=\sum_i d_i10^i \equiv \sum_i d_i(-1)^i \pmod{11}$, which is precisely the alternating digit sum $\sum_i (-1)^i d_i$.

Step 3. Conclude: $11\mid n$ exactly when $n\equiv0\pmod{11}$, which by Step 2 happens exactly when $11$ divides $\sum_i (-1)^i d_i$.