Prove that 11 divides a if and only if 11 divides the alternating sum of the digits of a

Question

Been stuck in this problem for quite a while. Apparently uses modular congruence to solve.

Answer 1

Write $a = \sum_{n=0}^N a_n10^n$ mod 11. Then as $10\equiv -1\mod 11$ we have

$$a\equiv\sum_{n=0}^Na_n(-1)^n\mod 11$$

which is exactly the alternating sum of its digits. If this is $0$, then by definition that's divisibility by $11$. This actually shows a stronger fact, not only is $a$ divisible by $11$ iff its alternating digital sum is, but in fact $a$ is always congruent to the alternating sum of its digits, be that sum $0$ or otherwise.

Answer 2

Hint:

$$10 \equiv -1 \mod 11$$

$$123=1 \times 10^2 + 2 \times 10 +3$$

Answer 3

Hint $ $ Radix notation has Polynomial form $\,n = d_0\! + d_1 10 + d_2 10^2\! +\cdots + d_k 10^k\! = P(10)\,$ so

${\rm mod}\ 11\!:\ \color{#c00}{10}\equiv \color{#c00}{-1}\,\Rightarrow\ n = P(\color{#c00}{10}) \equiv P(\color{#c00}{-1}) \equiv d_0 - d_1 + d_2 - \cdots + (-1)^k d_k\, $ by applying the Polynomial Congruence Rule, i.e. $\,a\equiv b\,\Rightarrow\,P(a)\equiv P(b)$.