Show the rule of divisibility for 11

Repeating the process on place values of thousands place, or more reviews the same rule. For even places, the remainders are the opposite numbers of the digits on the place and for odd places, the remainders are the same as the digits on the place, if divided by 11.

Next, let’s first 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 110b-11b is divisible by 11. And the remainder is b

For the place value on thousands place,

where the first part 1100a-110a+11a is divisible by 11 and the remainder is -a.

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

\begin{aligned} \overline{abcd}&=[(1100a-110a+11a)+(110b-11b)+ 11c]-a+b-c+d \end{aligned}

which shows the divisibility of the number by 11 is dependent on the difference of the sums of alternating places. If the difference is divisible by 11 or equal to 0, the number is divisible by 11.

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

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

The observation above reviews

and further,

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

If k=2n,

if k =2n-1

Then,

\begin{aligned} a &\equiv a_{2n}10^{2n}+a_{2n-1}10^{2n-1}+\dots+a_210^2+a_{1}10^1+a_0 \quad(\text{mod } 11) \\ &\equiv a_{2n}-a_{2n-1}+\dots+a_2-a_1+a_0 \quad (\text{mod } 11) \\ &\equiv (a_{2n}+a_{2n-2}+\dots+a_a+a_0) - (a_{2n-1}+a_{2n-3}+\dots+a_3+a_1) \quad (\text{mod } 11) \end{aligned}

Therefore, a is divisible by 11 if and only if the difference of the sums of digits on the even places and odd places is zero or divisible by 11.