# 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

## 4 Answers

\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}

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 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 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.)