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

Source Work

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

The Dictionary

$495$

Mistake

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:

\(\ds 191\)

\(\to\)

\(\ds 911 - 119\)

\(\ds \)

\(\to\)

\(\ds 792\)

\(\ds \)

\(\to\)

\(\ds 972 - 279\)

\(\ds \)

\(\to\)

\(\ds 693\)

\(\ds \)

\(\to\)

\(\ds 963 - 369\)

\(\ds \)

\(\to\)

\(\ds 594\)

\(\ds \)

\(\to\)

\(\ds 954 - 459\)

\(\ds \)

\(\to\)

\(\ds 495\)

Sources

• 1997: David Wells: Curious and Interesting Numbers (2nd ed.) ... (previous) ... (next): $495$