@quacker88 Actually there is a mistake in my proof. I will edit it later.
Question #1 isn't very hard.
S1: Suppose we have a number \(\overline{a_na_{n-1}\cdots a_2a_1a_0}\) where the \(a_i\)'s are digits. Writing this as a sum of powers of \(10\), we get
$$\overline{a_na_{n-1}\cdots a_2a_1}=a_n\cdot 10^n+a_{n-1}\cdot 10^{n-1}+\cdots+a_2\cdot 10^2+a_1\cdot 10+a_0.
$$
Notice that \(11\equiv -1\pmod{10}\); this means that \(11^n\equiv (-1)^n\pmod{10}\). So, we can rewrite the expression mod 9 as
$$a_n\cdot 10^n+a_{n-1}\cdot 10^{n-1}+\cdots+a_2\cdot 10^2+a_1\cdot 10+a_0\equiv a_n\cdot (-1)^n+a_{n-1}\cdot (-1)^{n-1}+\cdots+a_2\cdot (-1)^2+a_1\cdot (-1)+a_0\pmod{11}
$$
This is the alternating sum of digits, hence, we have proved the divisibility rule for 11.
Note 1. The number \(\overline{a_na_{n-1}\cdots a_2a_1a_0}\) has \(\boldsymbol{n+1}\) digits, not \(n\) digits.
Note 2.
If \(n\) is odd, then we will have the sum \(-a_n+a_{n-1}-a_{n-2}+\cdots+a_2-a_1+a_0\). This means we have to first negate the first digit and then do the alternating sum to find the remainder mod 11.
If \(n\) is even, then the sum is \(a_n-a_{n-1}+a_{n-2}-\cdots+a_2-a_1+a_0\). This means we leave the first digit as is and then do the alternating sum to find the remainder mod 11.
I will post my solution to Question #2 later.
