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2007-03-17 22:04:15 · 3 answers · asked by Sgeo 2 in Science & Mathematics Mathematics

Or, if it's too many, how many digits are there in the number of digits?

2007-03-19 14:24:02 · update #1

3 answers

There are ceiling(logn) digits in n, so there are:

ceiling[(2^65536)*log2] digits in 2^(2^65536),
where ceiling(x) is the next largest integer than x.

2007-03-17 22:11:30 · answer #1 · answered by Phineas Bogg 6 · 1 0

log(2^(2^65536)))
= log( 2.003529 930406 846464 979072 3515603 *10^19728)
= 19728 (0 d.p.)

2007-03-17 22:16:22 · answer #2 · answered by math freak 3 · 0 0

dang you guys are prodigies...anyway....MY ANSWER is....

a whole lot x[

2007-03-18 07:22:34 · answer #3 · answered by Lola Bunny 1 · 0 0

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