# Number is congruent to alternating sum of digits mod 11

Given that $a=a_n a_{n-1} ... a_1 a_0$, then $a=\sum_{i=0}^n a_i \cdot 10^i \equiv \sum_{i=0}^n a_i \cdot (-1)^i \mod{11}$, since $10 \equiv -1 \mod{11}$.