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Using the definitions at Wikipedia; Sloane; and Mathworld; I can't see why $1$ is a member of the Kaprekar series?

Would someone give an easy explanation?

Thanks.

(Yet more on this here).

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    Thanks DJC for the restoration of the links. I'm new so the spam filter (quite correctly) limited the number of links I could create).2011-06-08
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    **Found what I was after**: [The Kaprekar numbers](http://pictor.math.uqam.ca/~plouffe/OEIS/jis/The%20Kaprekar%20Numbers.pdf) by D.E.Iannucci says 1 is included by fiat!2011-06-08
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    I don't see what an Italian auto manufacturer has to do with it. $1$ is included because it satisfies the definitions, as both answerers agree.2011-06-08

2 Answers 2

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$1=0+1$, $1^2=0\times10^m+1$ seems to fit the definition as given at the OEIS reference.

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    It also fits the definition given at Wikipedia.2011-06-08
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    If we set m to its minimum i.e. 1; q is `0`, so r is 1. Then r is **not** `<(10^1)` ?2011-06-08
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    @Paddy3118, I don't understand. $10^1=10$, so, if $r=1$, then certainly $r\lt10^1$.2011-06-08
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    Umm, your right. But I do like the Iannucci paper just stating it is 'cos it is :-)2011-06-08
1

Unfortunately, my answer is anticlimactic. I happen to know wikipedia's definition (linked in the question, but I reproduce the definition here)

Let X be a non-negative integer. X is a Kaprekar number for base b if there exist non-negative integers n, A, and positive number B satisfying: $X^2 = Ab^n + B$, where $0 < B < b^n$, and s.t. $X = A + B$

So $A$ can be $0$. Thus $1^2 = 1 = 0* 10^1 + 1$, and we see that it's a Kaprekar number.

And - Gerry posted his answer just before me (I refreshed, and it's there)! But I wrote this too, so I'll keep it -

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    There is no need to sign your name at the end, since it appears at the bottom anyway.2011-06-08
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    Unfortunately the WP article goes on to state the `02011-06-08
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    @Paddy - this doesn't really matter, as it could just be 2 instead. Then we still see that 1 is such an example, as A is 0. That's the important part.2011-06-08
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    Yep. I now see where I was wrong. Thanks.2011-06-08