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Kaprekar discovered the Kaprekar constant or $6174$ in $1949$. He showed that $6174$ is reached in the limit as one repeatedly subtracts the highest and lowest numbers that can be constructed from a set of four digits that are not all identical.

e.g. starting with $1234$, we have $4321 − 1234$ = $3087$, then $8730 − 0378$ = $8352$, and $8532 − 2358$ = $6174$.

But, Why we reach to $6174$ through this process ? I think, subtraction is always divisible by $3$....(not sure)

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    See also http://oeis.org/A099009 and references there.2012-08-12

3 Answers 3

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$6174$ is a fixed-point of this process, i.e. $7641 - 1467 = 6174$. It turns out that it is the only fixed point, and there are no nontrivial cycles.

The sum of digits of the difference could also be $27$, e.g. for $6555-5556$.

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    Thanks for the comments. I missed the link to the paper and I now I see that this is a well-studied problem. The uniqueness method you suggest is pretty brute-force; I suppose this indicates that this is a pretty "artificial" question, without interesting mathematical content behind it.2013-01-30
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After one subtraction you will have a number divisible by $9$ because the remainder on division by $9$ is not changed when you rearrange the digits. That is not a problem as $6174$ is divisible by $9$. The statement that we always reach $6174$ rests on (as best I know) a search of the possibilities. If you just look at multiples of $9$ there are only $1000$ to look at, which is not so many even by hand. You only have to look at one permutation of each set of digits, which reduces the search considerably. There may be other ways to limit the search.

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When considering only 4 digit numbers, this problem can be expanded to any base and when you do that it can then be reduced to a system of 14 equations. 6174 is the only solution in base 10.

Other length 4 solutions are: $0111_2, 1001_2,3021_4, \qquad 3032_5, 6174_{10}, 92b6_{15}, c3f8_{20},\dots $