Kaprekar's Constant for three-digit numbers

Suppose we take any three-digit number, say x y z (I mean the number actually is 100x+10y+z).

we even consider that x>y>z (for ease of calculations).

So, the greatest and least possible numbers are (100x+10y+z) and (100z+10y+x) respectively.

Now, (100x+10y+z) - (100z+10y+x)
= 99(x-z)

now, we made an assumption that k=x-z, thus:

99k
= 100k - k
= 100k - 100 + 100 - 10 + 10 - k
= 100(k-1) + 90 + (10-k)

since k is an integer(or more precisely, a one digit number), (10-k) is also an integer. Thus when 99 is multiplied by k, the resultant number looks like (k-1) 9 (10-k).
we assume (k-1)<(10-k) So, if we decide to continue likewise, then: [900+10(10-k)+(k-1)] - [100(k-1)+10(10-k)+9] = 891 - 99(k-1) = 99(9-k+1) = 99(10-k) So, by our assumption, if we need to obtain a constant, then the subtraction result must be the same multiple of 99 we obtained a step ago. in that case, (10-k)=k which gives k=5. So, the desired Kaprekar's constant for three-digit numbers is 99x5=495. This process also claims that such a constant is unique for three-digit numbers.