495 (four hundred [and] ninety-five) is the natural number following 494 and preceding 496. It is a pentatope number[1] (and so a binomial coefficient).
The Kaprekar's routine algorithm is defined as follows for three-digit numbers:
Take any three-digit number, other than repdigits such as 111. Leading zeros are allowed.
Arrange the digits in descending and then in ascending order to get two three-digit numbers, adding leading zeros if necessary.
Subtract the smaller number from the bigger number.
Go back to step 2 and repeat.
Repeating this process will always reach 495 in a few steps. Once 495 is reached, the process stops because 954 – 459 = 495.
Example[]
For example, choose 495:
495
The only three-digit numbers for which this function does not work are repdigits such as 111, which give the answer 0 after a single iteration. All other three-digits numbers work if leading zeros are used to keep the number of digits at 3:
211 – 112 = 099
990 – 099 = 891 (rather than 99 - 99 = 0)
981 – 189 = 792
972 – 279 = 693
963 – 369 = 594
954 − 459 = 495
The number 6174 has the same property for the four-digit numbers.
See also[]
Collatz conjecture — sequence of unarranged-digit numbers always ends with the number 1.
References[]
^"Sloane's A000332". The On-Line Encyclopedia of Integer Sequences. OEIS Foundation. Retrieved 2016-05-16.
Eldridge, Klaus E.; Sagong, Seok (February 1988). "The Determination of Kaprekar Convergence and Loop Convergence of All Three-Digit Numbers". The American Mathematical Monthly. The American Mathematical Monthly, Vol. 95, No. 2. 95 (2): 105–112. doi:10.2307/2323062. JSTOR2323062.