Show the rule of divisibility for 9

Repeating the process on place values of thousands place, or more reviews the same rule. Division of place values by 9 give remainders that are the same as the digits on the places.

Next, let’s evaluate a 4-digit number that is denoted as \overline{abcd}, in which a, b, c, d are four digits on thousands, hundreds, tens and ones places respectively. Since a number can be expressed as the sum of its place values, \overline{abcd} is rewritten as

For place value on tens place,

For the place value on hundreds place,

where the first part 90b+9b is divisible by 9. And the remainder is b

For the place value on thousands place,

where the first part 900a+90a+9a is divisible by 91 and the remainder is a.

Substituting (2),(3) and (4) into (1) gives

\begin{aligned} \overline{abcd}&=[(900a+90a+9a)+(90b+9b)+ 9c]+a+b+c+d \end{aligned}

which shows the divisibility of the number by 9 is dependent on the sum of digits on all places of the number. If the sum is divisible by 9, the number is divisible by 9.

In a more general form, an integer can be expressed as

a=a_{n}10^{n}+a_{n-1}10^{n-1}+\dots+a_210^2+a_{1}10^1+a_0

The observation above reviews

and further,

where k = 1, 2,3,4 ...

Then,

\begin{aligned} a &\equiv a_{n}10^{n}+a_{n-1}10^{n-1}+\dots+a_210^2+a_{1}10^1+a_0 \quad(\text{mod } 9) \\ &\equiv a_{n}+a_{n-1}+\dots+a_2+a_1+a_0 \quad (\text{mod } 9) \\ \end{aligned}

Therefore, the number a is divisible by 9 if and only if the sum of digits on the all places of the number is divisible by 9.