Question

Let n = n_6n_5n_4n_3n_2n_1n_0 for some decimal number n. Prove that 11 | n if and if 11 |(n_0 - n_1 + n_2 - n_3 + n_4 - n_5 + n_6). Suggest an error detection procedure based on 3(c) above.

Answer 1

Here let us assume n is (breaking the digits of number)
n = x0 + 10 * x1 + 100 * x2 + … + 10kxk.
and another way of sum of digits of n is xk-xk-1+xk-2-.....+(-1)kx0
Now consider the expression :

E = (10 + 1) * (x0 + 10(x1 – x0) + 100(x2 – x1 – x0) + … + 10k(xk – xk-1 + xk-2 - …+ (-1)kx0)) – 10k+1(xk – xk-1 + xk-2 - … + (-1)kx0).
Let us consider the following expression:

After we perform multiplication to above we get E =n.
Now a term which is having a factor (10+1) i.e 11 is divisble by 11. Hence n is divisible 11 when we have 10k+1(xk – xk-1 + xk-2 - … + (-1)kx0)
Now 10K+1 will be divisble by 11 and 11 is prime hence n is divisble by 11 if and only if xk – xk-1 + xk-2 - … + (-1)kx0 is divisble by 11 and this is the alternate way of sum of the digits of n