Take any three-digit positive integer, reverse its digits, and subtract. For example, 742 − 247 = 495. The difference is divisible by 11. Prove that this must happen for all three-digit numbers abc
A more fun trick is:
Let $abc > cba$ (in other words) $a > c$.
Let $abc - cba = def$ (If $d = 0$ write it as a distinct zero)
Then $def + fed = 1089$.
To subtract $abc - cba$ you must subtract $a$ from $c$. But $a > c$ so you must borrow. That leaves $(10 + c - a)$ in the one's place. The ten's place is $b - b$ but you borrowed from the $b$ so you have $(b-1) - b$. $b > (b-1)$ so we must borrow again. So that leaves $10 + (b -1) + b = 9$ in the tens place. Then to subtract the hundreds place we have $(a-1) -c = (a-c - 1)$.
So $abc -cba = [a-c -1]9[10 + c-a]$
Reverse and add $[a-c -1]9[10 + c-a]+ [10+c-a]9[a-c-1]$
The ones: $(10 + c -a) + (a-c-1) = 9$.
The tens: $9+9 = 18$ write the $8$ and carry the $1$.
The hundreds $[a-c -1] + [10 + c -a]=9$ plus the $1$ we carried gives us $10$.
So $[a-c -1]9[10 + c-a]+ [10+c-a]9[a-c-1]=1089$ (which can be though of as $9*100 + 2*9*10 + 9$ which is $9(121) = 9*11^2$. (Food for thought.)
Also as $[a-c -1]9[10 + c-a]$ if you add the odd places $[a-c-1], [10+c-a]$ you get $9$. And if you add the even places you get $9$. As they are equal this is divisible by $11$. (And that explains why that rule works in the first place.)