How is 1 a member of the Kaprekar series?

Question

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).

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). — Paddy3118, Jun 08 '11 at 05:51
Found what I was after: The Kaprekar numbers by D.E.Iannucci says 1 is included by fiat! — Paddy3118, Jun 08 '11 at 06:32
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. — Gerry Myerson, Jun 08 '11 at 06:47

score 2 · Accepted Answer · answered Jun 08 '11 at 00:31

2

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

answered Jun 08 '11 at 00:31

Gerry Myerson

179,216

It also fits the definition given at Wikipedia. – JavaMan Jun 08 '11 at 00:32
If we set m to its minimum i.e. 1; q is 0, so r is 1. Then r is not <(10^1) ? – Paddy3118 Jun 08 '11 at 06:11
@Paddy3118, I don't understand. $10^1=10$, so, if $r=1$, then certainly $r\lt10^1$. – Gerry Myerson Jun 08 '11 at 06:44
Umm, your right. But I do like the Iannucci paper just stating it is 'cos it is :-) – Paddy3118 Jun 08 '11 at 06:55

davidlowryduda · Answer 2 · 2011-06-08T00:51:51.427

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 -

edited Jun 08 '11 at 00:51

answered Jun 08 '11 at 00:36

davidlowryduda

91,687

1

There is no need to sign your name at the end, since it appears at the bottom anyway. – Eric Naslund Jun 08 '11 at 00:42
Unfortunately the WP article goes on to state the 0<B<N with the minimum N being 1 so B cannot be 1 ? (the range on OEIS is slightly different - and in the WP nomenclature becomes 0<=B<N). – Paddy3118 Jun 08 '11 at 06:02
@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. – davidlowryduda Jun 08 '11 at 06:12
Yep. I now see where I was wrong. Thanks. – Paddy3118 Jun 08 '11 at 06:57

How is 1 a member of the Kaprekar series?

2 Answers2