Hi Mark,
There's no such number.
If S is one skwiirol, then:
S = 10^S ; S is a positive integer
LOG10(S) = S
No solution because LOG10(S) is always < S.
The nearest is S = 1 and LOG10(S) = 0.
(If S is a real number, S = 1/LN(10) = 0.4343
and LOG10(S) = -0.3622)
Bye,
Suyono
Mark wrote:
There's no such number.
If S is one skwiirol, then:
S = 10^S ; S is a positive integer
LOG10(S) = S
No solution because LOG10(S) is always < S.
The nearest is S = 1 and LOG10(S) = 0.
(If S is a real number, S = 1/LN(10) = 0.4343
and LOG10(S) = -0.3622)
Bye,
Suyono
Mark wrote:
>
> In the spirit of "googol" I've made up my own number. I've called it
> a "skwiirol" which is a really good name.
>
> It's defined as "1 followed by a skwiirol of zeroes". The best thing
> about it is that it is recursive - the more you try to work out how
> many zeroes it actually has, the bigger it gets!
>
> Mark
>