Is there an easy to solve this problem? I can find the answer by using a complicated rule that I don't understand. Even if I try to remember this rule, I probably will forget about it a year later. The rule is "To find out if a number is divisible by 11, add every other digit and call that sum $x$. Add together the remaining digits, and call that sum $y$. Take the positive difference of $x$ and $y$. If the difference is zero or a multiple of eleven, then the original number is a multiple of eleven."
A and B are non-zero digits for which A468B05 is divisible by 11. What is A+B?
$$x=4+8+0$$
$$y=A+6+B+5$$
Case 1: $x-y=0$
$$12-11-(A+B)=0$$
$$A+B=1$$
This means either $A$ or $B$ is $0$, so it doesn't work.
Case 2: $x-y=-11$
$$12-11-(A+B)=-11$$
$$A+B=12$$
$A$ and $B$ can be several combinations of positive nonzero integers that satisfies $A+B=12$, so this is the correct answer.