In article <[email protected]>,
 Steven D'Aprano <[email protected]> wrote:

> For what it's worth, that last intermediate result (two to the power of 
> the 489-digit number) has approximately a billion trillion trillion 
> trillion trillion trillion trillion trillion trillion trillion trillion 
> trillion trillion trillion trillion trillion trillion trillion trillion 
> trillion trillion trillion trillion trillion trillion trillion trillion 
> trillion trillion trillion trillion trillion trillion trillion trillion 
> trillion trillion trillion trillion trillion trillion digits.
> 
> (American billion and trillion, 10**9 and 10**12 respectively.)

What's that in crore?
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