I will try it. A question, though: is this only for certain numbers? Look at 22 (2*11):
Math.floor(sqrt(22))=4 factors of 4: 2 no 11? Have a great day, Alex New email address: mehgcap@xxxxxxxxx----- Original Message ----- From: "qubit" <lauraeaves@xxxxxxxxx>
To: <programmingblind@xxxxxxxxxxxxx> Sent: Friday, October 09, 2009 11:41 PM Subject: Re: prime factors shortcut question
take the floor of the square root. --le----- Original Message ----- From: "Alex Hall" <mehgcap@xxxxxxxxx>To: "Blind Programming List" <programmingblind@xxxxxxxxxxxxx> Sent: Friday, October 09, 2009 2:13 PM Subject: prime factors shortcut question Hi all,I am trying to find the prime factors of a massive number, over 600 billion,in perl (it is just an activity in a class, not an assignment). I know you can factor the square root, which is good since I cannot store such a bignumber in a simple scalar in perl, but I run into a problem: the square rootof my number is not a whole number, but a decimal. Of course, then, any modulus I try will not work since a decimal mod any whole number will not return 0. What do I do? Do I get the floor or ceiling of the square root operation? Is there another trick? I do not need something vastly complex that works for most situations or explains things from a cryptographicstandpoint, all I need is what to do with my decimal square root so moddingworks and gives me factors. Thanks for any help! Have a great day, Alex New email address: mehgcap@xxxxxxxxx __________ View the list's information and change your settings at //www.freelists.org/list/programmingblind __________ View the list's information and change your settings at //www.freelists.org/list/programmingblind
__________View the list's information and change your settings at //www.freelists.org/list/programmingblind