some divisibility tests

We want to derive the divisibility tests for 3, 9, and 11.

Let

$$A=a_n\cdot {10}^n+a_{n-1}\cdot {10}^{n-1}+\mathrm{\dots }+a_0$$

be the digital representation of the integer $A$ in the decadic system ($0\le a_i\le 9$).

Since  $10\equiv 1(mod3,\mathrm{\hspace{0.17em}9})$,  we have also  ${10}^i\equiv 1(mod3,\mathrm{\hspace{0.17em}9})$  for each $i$,  and hence

$$a_i\cdot {10}^i\equiv a_i(mod3,\mathrm{\hspace{0.17em}9}).$$

Consequently

$$A=\sum _{i=0}^n{a}_i\cdot {10}^i\equiv \sum _{i=0}^n{a}_i(mod3,\mathrm{\hspace{0.17em}9}).$$

This congruence means that

${\displaystyle \sum _{i=0}^n}{a}_i\equiv \mathrm{\hspace{0.33em}0}(mod3,\mathrm{\hspace{0.17em}9})\mathit{\hspace{1em}\hspace{1em}}\iff \mathit{\hspace{1em}\hspace{1em}}A\equiv \mathrm{\hspace{0.33em}0}(mod3,\mathrm{\hspace{0.17em}9}).$(1)

The fact  $10\equiv -1(mod11)$  similarly yields

$$A=\sum _{i=0}^n{a}_i\cdot {10}^i\equiv \sum _{i=0}^n{(-1)}^i{a}_i(mod11),$$

whence

${\displaystyle \sum _{i=0}^n}{(-1)}^i{a}_i\equiv \mathrm{\hspace{0.33em}0}(mod11)\mathit{\hspace{1em}\hspace{1em}}\iff \mathit{\hspace{1em}\hspace{1em}}A\equiv \mathrm{\hspace{0.33em}0}(mod11).$(2)

The results (1) and (2) may be rendered as the following:

• $A$ is divisible by 3 (resp. 9) if and only if  $a_0+a_1+\mathrm{\dots }+a_n$  is.
  

• $A$ is divisible by 11 if and only if  $a_0-a_1+-\mathrm{\dots }+{(-1)}^n{a}_n$  is.