let s = $a_{0}+a_{1}+......+a_{n}$
let k = $a_{0}+a_{1}*10^{1}+......+a_{n}*10^{n}$
k-s = $a_{1}*(9_{1})+a_{2}*(9_{1}9_{2})+......+a_{n}*(9_{1}9_{2}....9_{n})$
It is obvious that 11 will divide any term in the sequence where the number of 9's is even. so we only have to worry about the odd terms
let k-s = t+r
where t contains terms with the even number 9's, which are all divisible by 11 and r contains the terms with odd numbers
r = $a_{1}*(9_{1})+a_{3}*(9_{1}9_{2}9_{3})+......a_{2n+1}*(9_{1}9_{2}....9_{2n+1})$
each term in r can be decomposed for all $(2n+1)>1$
$a_{2n+1}*(9_{1}9_{2}....9_{2n+1})= a_{2n+1}*(9*10^{2n}) + a_{2n+1}*(9_{1}....9_{2n})$
where $a_{2n+1}*(9_{1}....9_{2n})$ is divisible by 11
we know that $(9*10^{2n})$ is never divisible by 11. Therefore for r to be divisible by 11 all $a_{2n+1}$ must be divisible by 11.
Therefore for k-s to be divisible by 11 the sum of all the odd digits must equal to 11
k = (k-s) + s = t+r + s
using what we have already proved if k is divisible by 11 then t is divisible by 11, always, and the r is dividable by 11 if the sum of the odd terms of digits is 11
lastly the sum of the even terms must also be be divisible by 11 as s must be divisible by 11 for k to be divisible by 11
Therefore the sum of the odd and even terms in the digits must sum to a number divisible by 11 for K to be divisible by 11
P.S here is the correct proof
let k = $a_{0}+a_{1}*10^{1}+ a_{2}*10^2......+a_{n}*10^{n}$
CASE 1 ODD NUMBER OF DIGITS in K expansion:
k = $a_{0}+a_{1}*(1_{1}1_{2})-a_{1}+a_{2}*(9_{1}9_{2})+a_{2}......a_{n}*(1_{1}1_{2}....1_{n})-a_{n}$
k = $a_{1}*(1_{1}1_{2})+a_{2}*(9_{1}9_{2})+......+a_{n}*(1_{1}1_{2}....1_{n}) + (a_{0}-a_{1}+a_{2}-.......-a_{n})$
We can see that all the terms $a_{1}*(1_{1}1_{2})+a_{2}*(9_{1}9_{2})+......+a_{n}*(1_{1}1_{2}....1_{n})$ are divisible by 11
Therefore for k to be divisible by 11 the $(a_{0}-a_{1}+a_{2}-.......-a_{n})$ must divisible by 11 or equal zero. Which is the difference of the odd and even terms
CASE 2 : EVEN NUMBER OF DIGITS
WLOG we apply the same argument for an even number of digits. The only difference in the proof is that the last term in $(a_{0}-a_{1}+a_{2}-.......)$ will have a different sign. However the end result will mean the same. That the difference in odd and even digits determines whether a number is divisible by 11