Editing Divisibility rule (section)

====Case where all digits are summed====

This method works for divisors that are factors of 10&nbsp;−&nbsp;1 = 9.

Using 3 as an example, 3 divides 9&nbsp;=&nbsp;10&nbsp;−&nbsp;1. That means <math>10 \equiv 1 \pmod{3}</math> (see [[modular arithmetic]]). The same for all the higher powers of 10: <math>10^n \equiv 1^n \equiv 1 \pmod{3}.</math> They are all [[congruence relation|congruent]] to 1 modulo 3. Since two things that are congruent  modulo 3 are either both divisible by 3 or both not, we can interchange values that are congruent modulo 3. So, in a number such as the following, we can replace all the powers of 10 by 1:

: <math>100\cdot a + 10\cdot b + 1\cdot c \equiv (1)a + (1)b + (1)c \pmod{3},</math>

which is exactly the sum of the digits.