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 Post subject: 6174=Kaprekar constant
PostPosted: Fri, 10 Feb 2006 04:25:59 UTC 
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The number 6174 arises in the following famous problem :

Take any 4-digit number x_1 which uses more than one digit and find the difference x_2:=T(x_1) between the numbers formed by writing the digits in descending order and ascending order.
For example, starting with x_1= 1470 yields x_2=T(x_1)=7410- 0147=7263. Iterate this process using the difference x_2 as the new 4-digit number. In other words,
\begin{array}{lcl}
x_3&=&T(x_2)=7632-2367=5265 \\
x_4&=&T(x_3)=6552-2556=3996\\
x_5&=&T(x_4)=9963-3699=6264\\
x_5&=&T(x_5)=6642-2466=4176\\
x_6&=&T(x_6)=7641-1467={\mathbf 6174}\\
x_7&=&T({\mathbf 6174})={\mathbf 6174} 
\end{array}
[/etx]
The <strong>Indian mathematician  <span style="color: blue"> D.R. Kaprekar </span> </strong> discovered that this process leads in at most 7 steps to the number [tex] {\mathbf 6174}=[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 für 2-stellige g-adische Zahlen,
(German, Spanish summary) Rev. Real Acad. Cienc. Exact. Fís. 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 r-digit g-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:
PostPosted: Fri, 10 Feb 2006 11:54:45 UTC 
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Hi Alex,

Some observations on this problem.

Let \displaystyle a,\, b,\, c,\, d, be the digits of your number.
So the first sum is \displaystyle (1000a+100b+10c+d) - (1000d+100c+10b+a)

This gives us \displaystyle (999a-999d+90b-90c)
which we can rearrange slightly to give us \displaystyle (999a-999d)\,+\,(90b-90c)
.....A

Then we realise that \displaystyle a=d+x,\;b=d+x-1,\;c=d+x-2, (or similar)
so we re-write the terms like this: \displaystyle 999(d+x)= 999d + 999x

............................................\displaystyle 90(d+x-1) = 90d + 90x - 90
............................................\displaystyle 90(d+x-2) = 90d + 90x - 180

Re-write A using the terms in this form, and we get:
.....\displaystyle (999d + 999x - 999d) + (90d + 90x - 90 - 90d + 90x - 180)
which simplifies to \displaystyle 999x+90

Now, since \displaystyle x>2, x<10 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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PostPosted: Fri, 10 Feb 2006 13:01:16 UTC 
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Numbers wrote:
Hi Alex,....Some observations on this problem....

Thanks for interesting remarks !


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PostPosted: Fri, 10 Feb 2006 15:47:03 UTC 
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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 :idea:

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 Post subject:
PostPosted: Fri, 10 Feb 2006 18:33:28 UTC 
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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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