Quote:
Originally Posted by retina
Okay. But it still doesn't meet the criterion 99%.
log(6¹⁰⁰)/log(7×6¹⁰⁰+11) = 98.92...%
Indeed for any integer N I can't see how you ever get 99%. The only way would be something like e×e⁹⁹+0 = 99%. So k=e, b=e, n=99, c=0. And other multiples of those values.

Ah! You're thinking 99% as an exact number, whereas I'm pretty sure OP intended to give the general flavor of the structure of the number. Meaning, b^n part is the dominant term.