Consider the positive integer 11.

1 - 1 = 0, which is divisible by 11.

Let the statement be true for a positive integer m, which is divisible by 11.

$\displaystyle S(m)=a_1-a_2+a_3-...+(-1)^{n-1}a_n$ is divisible by 11.

Consider the positive integer m + 11.

If the last digit of m is <= 8, it will increase by 1. If i digits before the last digit are equal to 9, they will all become 0 in m + 11. The digit before the i digits (equal to 9) will increase by 1, with the result that S (m + 11) = S (m) if i is even and $\displaystyle S(m+11)=S(m)\pm9\pm2=S(m)\pm11$ if i is odd.

If the last digit of m is 9, the last digit of m + 11 becomes 0. If the second last digit of m is < 8, it increases by 2, with the result that $\displaystyle S(m+11)=S(m)\pm9\pm2=S(m)\pm11$. If, however, the second last digit of m is 8 or 9 it becomes 0 or 1 respectively. If i digits before the second last digit are equal to 9, they all become 0 and the digit before these i digits increases by 1, with the result that $\displaystyle S(m+11)= S(m)\pm9\mp8\mp1=S(m)$ if i is even and $\displaystyle S(m+11)=S(m)\pm9\mp8\pm9\pm1=S(m)\pm11$ if i is odd.

So, the sum of digits of n + 11 is also divisible by 11.

This completes the proof.

Could someone please check my proof and see if I have made any mistakes or missed out any cases? Thanks.