Lets see... 267573734103 = 12 Octal Digits = 12*3 bits 010 110 111 101 111 011 111 011 100 001 000 011 2 6 7 5 7 3 7 3 4 1 0 3
-----Original Message----- From: Edmund Cramp [mailto:[EMAIL PROTECTED] Sent: Wednesday, June 19, 2002 4:42 PM To: [email protected] Subject: RE: [brlug-general] Math/computer question > a Hex number is base 16. In binary this would take four bits to > represent. Such as the hex number "a" would be 1011. Therefore a hex > number of "05bde5b843" is 10 digits long. Therefore it would be a 40 > bit key right? Correct on all points - it's 40-bits - let's count them <grin> 0000,0101,1011,1101,1110,1111,1011,1000,0100,0011 > Just making sure that I am not crazy.... No more than the rest of us. Now for bonus bits lets try the exercise in Octal... 267573734103 -- Edmund Cramp http://www.emgsrus.com/graffiti.htm _______________________________________________ General mailing list [email protected] http://brlug.net/mailman/listinfo/general_brlug.net
