Kaprekar's Process on 3 Digit Number ends in 495/Mistake

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Source Work

1997: David Wells: Curious and Interesting Numbers (2nd ed.):

The Dictionary


Take any $3$-digit number whose digits are not all the same and is not a palindrome. Arrange its digits into ascending and descending order and subtract. Repeat. This is called Kaprekar's process. All $3$-digit numbers eventually end up with $495$.

The $3$-digit number in question needs only to have its digits not all the same. A palindrome ends up at $495$ in the same way as any other number, for example:

\(\ds 191\) \(\to\) \(\ds 911 - 119\)
\(\ds \) \(\to\) \(\ds 792\)
\(\ds \) \(\to\) \(\ds 972 - 279\)
\(\ds \) \(\to\) \(\ds 693\)
\(\ds \) \(\to\) \(\ds 963 - 369\)
\(\ds \) \(\to\) \(\ds 594\)
\(\ds \) \(\to\) \(\ds 954 - 459\)
\(\ds \) \(\to\) \(\ds 495\)