Divisibility by 11

In my previous post on divisibility, I had shared the logic applied behind the divisibility check for 3 and 9. Here, I share the working behind divisibility check by 11.

Divisibility by 11 : Method

A number is said to be divisible by 11, if the difference of sums of its alternate digits is divisible by 11.

It simply means,

Sum up the digits on odd place values and those on even place values separately => Find the difference between the sums => if the resulting value is divisible by 11, then the number is divisible by 11, as indicated below:

For N = abcdef,
sum1 = a + c + e
sum2 = b + d + f
calculate difference => sum1 – sum2
if difference value is divisible by 11, then N is divisible by 11
For example:-
N = 5832,
sum1 = 5 + 3 = 8
sum2 = 8 + 2 = 10
difference => sum1 – sum2 => -2
2 is not divisible by 11, thus N=5832 is not divisible by 11

Divisibility by 11 : Logic

Let’s see the logic and working behind why and how the method works!

For any N = abcdef, we rewrite it as:  
a x (105) + b x (104) + c x (103) + d x (102) + e x (101) + f x (100
which is same as: 
105a + 104b + 103c + 102d + 101e + 100f  
=>  105a + 104b + 103c + 102d + 10e + f 
We write 10 = 11 – 1, and we get… 
(11-1)5a + (11-1)4b + (11-1)3c + (11-1)2d + (11-1)e + f  
Using the distributive laws and assuming all the multiples of 11 as m, we write 
[11m + (-1)5]a +[11m + (-1)4]b + [11m + (-1)3]c + [11m + (-1)2]d + [11m + (-1)]e + f 
Factoring out the common term 11m, we get
N = 11m [a + b + c + d + e] – a + b – c + d – e + f

Here, we can say:

[- a + b – c + d – e + f] is same as (b + d + f) – (a + c + e), which is nothing but the difference between sums of alternate digits, as described in the method earlier.

Through the above equation, we see that for N to be divisible by 11, the term [(b + d + f) – (a + c + e)] needs to be divisible by 11.

Let’s see few examples:

N= 45831 
N = 4 x (104) + 5 x (103) + 8 x (102) + 3 x (101) + 1 x (100
= 4 x (11-1)4 + 5 x (11-1)3 + 8 x (11-1)2 + 3 x (11-1) + 1 
= 4 [11m + (-1)4] + 5 [11m + (-1)3] + 8 [11m + (-1)2] + 3 [11m + (-1)]d + 1 
= 11m [4 + 5 + 8 + 3] + 4 – 5 + 8 – 3 + 1 
= 11m [20] + 5 
= 11m + 5
5 is not divisible by 11, hence 45831 is not divisible by 11 
For N = 75933 
N = 7 x (104) + 5 x (103) + 9 x (102) + 3 x (101) + 3 x (100
= 7 x (11-1)4 + 5 x (11-1)3 + 9 x (11-1)2 + 3 x (11-1) + 3 
= 7 [11m + (-1)4] + 5 [11m + (-1)3] + 9 [11m + (-1)2] + 3 [11m + (-1)]d + 3  
= 11m [7 + 5 + 9 + 3] + 7 – 5 + 9 – 3 + 3 
= 11m [24] + 11 
= 11m + 11
Here, N = 45837 is divisible by 11 
See below a very simple example without the assumption of multiples of 11 as m.
For N = 121
= 1 x 102 + 2 x 10 + 1 x 1

= 1 [11-1]2 + 2 [11-1] + 1

= 1 [112 – 11(2) + 1] + 2[11 + (-1)] + 1

= 1 [11(11 -2) + 1] + 2 [11 + (-1)] + 1

= 1 [11(9) + 1] + 2 [11 + (-1)] + 1

= 11 [9+2] + 1 – 2 + 1

= 11[11] + 0
=> 0 is considered to be divisible by all numbers. Thus N=121 is seen as divisible by 11

2 Comments Add yours

  1. Nithya Veena says:

    Very Interesting! Thank you for sharing the logic and refreshing the facts.. keep sharing more maths logic 😊

    Liked by 1 person

    1. Aswini says:

      Thank you 🙂

      Like

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