Computing $\frac{|abcxyz-xyzabc|}{|abc-xyz|}=999$

Question

How do I compute:

$$\frac{|(x-a)\cdot 10^5 + (y-b)\cdot 10^4 + (z-c)\cdot 10^3 + (a-x)\cdot 10^2+ (b-y)\cdot 10 + (c-z)|}{|(x-a)\cdot 10^2 + (y-b)\cdot 10 + (z-c)|}$$

I'm stucked at this point. I tried to estimate with triangle inequality, but this doesn't seem to work.

The solution should be $999$.

It's from a math riddle where you have two numbers $abc$ and $xyz$ and $abc\neq xyz$.

When you concatenate them to get $abcxyz$ and $xyzabc$ (not multiplication) you can calculate $$\frac{|abcxyz-xyzabc|}{|abc-xyz|}=999$$

Answer 1

It should be written $abc\neq xyz,$ otherwise doneminator is zero.

You can compute by writing $abcxyz=1000(abc)+xyz$ then

$$\dfrac {|1000(abc)+xyz-1000(xyz)-abc|}{|abc-xyz|}=999\dfrac{|abc-xyz|}{|abc-xyz|}=999$$