How many digits are there in 2^1000 (2 to the power of 1000)?
(In reply to
check this out by akila)
Except that the answer is 302 digitsnot 301. The technique of "for every increase of 10 powers resultant answer increases by 3 digits" actually is assuming that 2^10 = 1000. It does not. It equals 1024. Thus 1K of memory has 1024 bytes. 1 Meg has 1,048,576. Eventually the excess leads to an extra digit, and this happens every so often. For example, 2^93 has 28 digits, but 2^103 has 32; 2^186 has 56 digits, but 2^196 has 60. Referring back to the log(2) method previously posted, the number of digits in 2^n is one more than the integer part of n*log(2). The method used by akila assumes that log(2) = .3, when in fact it is .3010299956639811.
P.S. Anecdote: an approximation of log(2) became imbedded in my head many years ago when an entrance to Columbia University as 3010 Broadway had its street address preceded by a graffito "log(2) = ."

Posted by Charlie
on 20030302 10:52:57 