Divisibility by11
Proof:
Let the number be abcde.
Remainderabcde 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. 10^{x} 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)×(111) = (sum)×11sum 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.
Examples
