ah, the old "odd - even" question.
in all of mathematics, if a number is evenly divisible by "2" ( with no remainder" then it is an even number. by definition. in binary, the extension of this principal is that if the right most bit is "0" then the number is even. you could count it out on your fingers or on a piece of paper, but that's the way it works. I.e. any binary number whose rightmost bit is zero will be some value that is divisible evenly by "2" recall that the binary values are 128 64 32 16 8 4 2 and 0 a binary number can only be odd if that last bit is a "1" boring, I know, but this gets you to the good part. set your "care" bit so that the rightmost bit "must" be zero, and you guarantee that the value will be even. set the "care" bit so that the rightmost bit is "1" and you guarantee that the value will be odd. make sense? -- www.chuckslongroad.info like my web site? take the survey! ""Cisco Nuts"" wrote in message [EMAIL PROTECTED]">news:[EMAIL PROTECTED]... > Hello, > > I am trying to understand how a wildcard mask of 0.0.254.255 filters odd > routes so that only even routes get across the router. > > Ex. If you have routes for 172.168.1.0/24, 2.0/24, 3.0/24, 4.0/24, 5.0/24, > 6.0/24 and you have an access-list of: > > #access-list 11 deny 172.168.1.0 0.0.254.255 > #access-list 11 permit any > > And a: > > #distribute-list 11 out > > This will allow only: 172.168.2.0/24, 4.0/24 and 6.0/24 routes out. > > Anyone can kindly explain the magic of this 0.0.254.255 wildcard? > > Thank you. > > Sincerely. > > > BTW: Anyone know the link on this on CCO? > > > > > > > > > _________________________________________________________________ > Send and receive Hotmail on your mobile device: http://mobile.msn.com Message Posted at: http://www.groupstudy.com/form/read.php?f=7&i=54498&t=54496 -------------------------------------------------- FAQ, list archives, and subscription info: http://www.groupstudy.com/list/cisco.html Report misconduct and Nondisclosure violations to [EMAIL PROTECTED]
