11

Divisibility by11

  • Method 1:
    1. Find sum of digits with place value of even powers of ten.
    2. Find sum of digits with place value of odd powers of ten and multiply by 10.
    3. Add the results of step 1 and 2.
    4. Repeat the process if the result is large.
    5. If result of step 4 is divisible by 11 then the number is divisible by 11.

  • Method 2: If the sum of digits at even positions form right and sum of digits at odd places from right has difference zero or a multiple of eleven then the number is divisible by eleven else not.

Proof:

Let the number be abcde.

abcde can be written as
a×10000 + b×1000 + c×100 + d×10 + e

which can also be written as
a×(909×11+1) + b×(90×11+10) + c×(9×11+1) + d×10 + e

which can also be written as
a×(909×11)+ a + b×(90×11)+ 10b + c×(9×11)+ c + d×10 + e

a×(909×11)+ b×(90×11) + c×(9×11) is divisible by 11 as they have 11 as common factor.

Every even power of 10 i.e. 10x when divided by 11 has remainder 1 and every odd power of 10 has remainder 10.

The even powers of ten occur at odd places from right and odd powers of ten occur at even places from right in the number.

So digits whose place value has odd powers of ten are added and multiplied by ten.
If this sum and sum of the digits whose place value has even powers of ten is added and is divisible by 11 then the number is divisible by eleven else not.

This can be represented as
(sum)×10
= (sum)×(11-1) = (sum)×11-sum
Here, (sum)×11 is divisible by 11 so the amount left is -sum.
This is added to the sum of digits with place value of even powers of 10.
If the result is a multiple of 11 or zero then the number is divisible else not.
Remainder

  • The remainder of the method 1 is the remainder of the original number.

  • In method 2: If negative number is got then the modulus of the number or (number + the negative value) can be the remainder.
    If positive number is got then the number or (number - positive number) can be the remainder.

Examples

  1. Is 348 divisible by 11.
    The sum by method 1 is (4)10+3+8 = 51. 7 is the remainder.
    The sum by method 2 is -3+4-8=-7. The number is not divisible.

  2. Is 3443 divisible by 11.
    Method 1: (3+4)10+3+4=77. The number is divisible.
    Method 2: Sum of digits is -3+4-4+3 = 0. The number is divisible.


Comments