Take any three-digit positive integer, reverse its digits, and subtract. The difference is divisible by 11. 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: \begin{align}
x &= 100a + 10b + c \quad &\text{Any three digit number can be expressed in this form}\\
y &= 100c + 10b +a \quad &\text{This is the number once the digits are reversed}\\
x-y &= 100a - a + 10b -10b +c -100c = 99a-99c\\
x-y &=11 \cdot(9a-9c)
\end{align}
A: The first number, what you've called "$abc$", is equal to $100a+10b+c$. "$cba$" is $100c+10b+a$. Thus, their difference is:
$$100a+10b+c - (100c+10b+a)\\
=99a-99c\\
=11(9a-9c).$$
Since $9a-9c$ is necessarily an integer, $11$ divides $abc-cba$.
A: A three digit number can be expressed as
$100x+10y+z \cdots1$
Where $x,y,z$ are the digits in the number .
Lets reverse it 
We get 
$100z+10y+x \cdots2 $
Subtract equation 2 from 1
We get 
$99x-99z$
Which can be expressed as 
$11(9x-9z)$
Thus it will be multiple of 11.
A: 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.)
