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 Post subject: 6174=Kaprekar constantPosted: Fri, 10 Feb 2006 04:25:59 UTC
 Member of the 'S.O.S. Math' Hall of Fame

Joined: Thu, 26 Jan 2006 18:51:28 UTC
Posts: 345
Location: Sibiu ,Romania
The number 6174 arises in the following famous problem :

Take any 4-digit number which uses more than one digit and find the difference between the numbers formed by writing the digits in descending order and ascending order.
For example, starting with yields . Iterate this process using the difference as the new 4-digit number. In other words,
[i ]Kaprekar's constant[/i], a fixed point of the iteration.

QUESTION: Justify the Kaprekar's routine. Try to find generalizations of this algorithm .

Remarks:
1) Regarding Kaprekar's original publications: He self-published most of his results, via a small Indian publishing company at his own expense.
2) It seems that by using Kaprekar's routine , exactly 77 four-digit numbers, namely 1000, 1011, 1101, 1110, 1111, 1112, 1121, 1211, ...
(see [13] Sloane's A069746), reach 0, while the remainder give 6174 in at most 8 iterations.

REFERENCES:
[1] Deutsch D. and Goldman B. ,Kaprekar's Constant, Math. Teacher 98,(2004) 234--242.
[2] Eldridge, K. E. and Sagong, S. The Determination of Kaprekar Convergence and Loop Convergence of All 3-Digit Numbers,
Amer. Math. Monthly 95, (1988) 105--112.
[3] Furno A.L. , J. Number Theory 13, no.2, (1981) 255--261.
[4] Hasse H. , Iterierter Differenzbetrag fÃ¼r -stellige -adische Zahlen,
(German, Spanish summary) Rev. Real Acad. Cienc. Exact. FÃ­s. Natur. Madrid 72 (1978), no. 2, 221--240.
[5] Hasse H. and Prichett G. D. , The determination of all four-digit Kaprekar constants,
J. Reine Angew. Math. 299/300 (1978), 113--124.
[6] Kaprekar D. R. , An Interesting Property of the Number 6174, Scripta Math. 15,(1955) 244--245.
[7] Kiyoshi Iseki , Note on Kaprekar's constant, Math. Japon. 29 (1984), no. 2, 237--239.
[8] Lapenta J. F., Ludington A. L. and Prichett G. D., An algorithm to determine self-producing -digit -adic integers, J. Reine Angew. Math. 310 (1979) 100--110.
[9] Ludington Anne L., A bound on Kaprekar constants, J. Reine Angew. Math. 310 (1979), 196--203.
[10] Prichett G. D., Ludington A. L. and Lapenta J. F., The determination of all decadic Kaprekar constants, Fibonacci Quart. 19 , no. 1,(1981) 45--52.
[11] Rosen Ken, Elementary Number Theory Book, (4-th. ed, see page 47)
[12] A. Rosenfeld, Scripta Mathematica 15 (1949), 241--246,
[13] Sloane, N. J. A. Sequences A069746,A090429, A099009, and A099010 , in The On-Line Encyclopedia of Integer Sequences,
http://www.research.att.com/~njas/sequences/.
[14] Trigg, C. W. , All Three-Digit Integers Lead to..., The Math. Teacher, 67 (1974) 41-45.
[15] Young, A. L. , A Variation on the 2-digit Kaprekar Routine, Fibonacci Quart. 31,(1993) 138--145.

[ Perhaps of interest ,Alex]

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 Post subject: Posted: Fri, 10 Feb 2006 11:54:45 UTC
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Joined: Fri, 19 Aug 2005 17:30:54 UTC
Posts: 386
Location: Somewhere near Beetlegeuse
Hi Alex,

Some observations on this problem.

Let be the digits of your number.
So the first sum is

This gives us
which we can rearrange slightly to give us
.....A

Then we realise that , (or similar)
so we re-write the terms like this:

............................................
............................................

Re-write A using the terms in this form, and we get:
.....
which simplifies to

Now, since the answer to your sum can take only one of 7 different values;
3087, 4086, 5085, 6084, 7083, 8082, 9081. and from any of those, you must eventually
end up with 6174.

This does of course assume that a, b, c, d are unique. I haven't had time to look at what
happens when b = c, etc.

Hope this is useful,

Numbers

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There are lots of unsolved problems in mathematics, and many of them can make you rich. Unfortunately they all seem to be pretty darn hard...
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 Post subject: Posted: Fri, 10 Feb 2006 13:01:16 UTC
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Joined: Thu, 26 Jan 2006 18:51:28 UTC
Posts: 345
Location: Sibiu ,Romania
Numbers wrote:
Hi Alex,....Some observations on this problem....

Thanks for interesting remarks !

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 Post subject: Posted: Fri, 10 Feb 2006 15:47:03 UTC
 Member of the 'S.O.S. Math' Hall of Fame

Joined: Sun, 24 Jul 2005 20:12:39 UTC
Posts: 4930
Location: Ottawa Ontario
Just ran a short program: there are (as you say) 77 such 4digit numbers.
Of those we have 9 that are obvious: 1111, 2222, ... , 9999.

So there is 68 such numbers (not counting the above 9).

I'll contribute the following observations:

ALL of these 68 numbers have 3 digits the same, the other digit being 1 more or 1 less.

ALL of these 68 result in an initial difference of 999; so 1 iteration only.

The 1st 4 are 1000, 1011, 1101, 1110 (they screw up the following pattern!)

The remaining 64 are really only 16 numbers:

1112 : 1121, 1211, 2111
1222 : 2122, 2212, 2221
similarly (hate typing!):
2223
2333

3334
3444

4445
4555

5556
5666

6667
6777

7778
7888

8889
8999

Interesting? Not if it doesn't bring down the price og groceries

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I'm just an imagination of your figment...

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 Post subject: Posted: Fri, 10 Feb 2006 18:33:28 UTC
 Member of the 'S.O.S. Math' Hall of Fame

Joined: Thu, 26 Jan 2006 18:51:28 UTC
Posts: 345
Location: Sibiu ,Romania
Denis wrote:
there are (as you say) 77 such 4digit numbers. Of those we have 9 that are obvious: 1111, 2222, ... , 9999.

Very interesting, nice,Alex

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