So, I'm pretty sure it's common knowledge that 11 has some tricks when it comes to multiplying - most notably that you can add numbers together to get products. One example is $11 * 23$ which is $235$ since $2 + 3 = 5$, of course.
I've been trying to expand this to a general case and then write a proof for it. I'm pretty sure the general case is $$11x=(10^n)(a_1)+(10^{n-1})(a_1+a_2)+(10^{n-2})(a_2+a_3)+...+(10^1)(a_{n-1}+a_n)+(10^0)(a_n)$$
where $x$ is a number with $n$ digits such that $$x = (10^{n-1})(a_1)+(10^{n-2})(a_2)+...+(10^0)(a_n)$$
So then I was going to use PMI to solve it. I solved for the $P(1)$ and $P(2)$ cases, then moved on to where I had $P(k)$ as $$11[(10^{k-1})(a_1)+(10^{k-2})(a_2)+...+(10^0)(a_k)] = (10^k)(a_1)+(10^{k-1})(a_1+a_2)+(10^{k-2})(a_2+a_3)+...+(10^1)(a_{k-1}+a_k)+(10^0)(a_k)$$
From here, I had $P(k+1)$ as $$11[(10^{(k+1)-1})(a_1)+(10^{(k+1)-2})(a_2)+...+(10^0)(a_{k+1})] = (10^{k+1})(a_1)+(10^{(k+1)-1})(a_1+a_2)+(10^{(k+1)-2})(a_2+a_3)+...+(10^1)(a_{(k+1)-1}+a_{k+1})+(10^0)(a_{k+1})$$
This is unfortunately where I got stuck. Beyond simplifying the exponents, I don't remember what to do at this point and need some assistance. I thought about expanding, but I don't think that's right. I also thought about adding terms but I don't quite know if that's right either.