tag:blogger.com,1999:blog-9732402.post111114468235115739..comments2009-11-14T12:04:47.719-08:00Walter GRhttp://www.blogger.com/profile/04858350394137675339noreply@blogger.comBlogger2125tag:blogger.com,1999:blog-9732402.post-89308052293369117312009-11-14T12:04:47.719-08:002009-11-14T12:04:47.719-08:00Who knows where to download XRumer 5.0 Palladium? <br />Help, please. All recommend this program to effectively advertise on the Internet, this is the best program!Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-9732402.post-1111293567273972852005-03-19T20:39:00.000-08:002005-03-19T20:39:00.000-08:00Unfortunately, you've made an error in your formula here. You're making the assumption that meta is a variable being multiplied by blogger, in otherwords that<BR/>Metablogger = meta * blogger<BR/><BR/>Pretty clearly however, both a-list and meta are functions, i.e. metablogger <--> meta(blogger) == A-List(blogger). It seems then, that what we might want to figure out then, is...<BR/>A-List(A-List(blogger))...however, hence all we really know is that A-List(A-List(blogger)) == A-List(meta(blogger)). Which does imply that in order to be in the A-List of the A-List, once also needs to be a metablogger. Unfortunately, I don't know enough about the properties of either function to take this further. Which is probably why I'm not currently a member of A-List(Blogger)Derek Wollensteinhttp://www.blogger.com/profile/05218257648142986346noreply@blogger.com