wildassmasks...

wild card masks...  i've linked this before but the site below is truly invaluable...

http://www.dslreports.com/faq/15216

my take on it...

128 192 224 240 248 252 254 256
1      1     1     1    1     1     1     1

count em up...

255.255.255.192 = (action in the fourth octet) /26 or 24 + 2 one's
255.255.240.0 = (action in the third) /20 or 16 + 4 one's

ok, we love 256...  256 - 192 = (in the first example) 64
64 minus 1 in the fourth octet =
0.0.0.63  is the wildass

second example; 256 - 240 = 16 -1 or 15 in the third is
0.0.15.255 is the wildass
it's easier to subtract from 256 than 255, and faster... wild... 

summarization redux...

are you sick of writing down all the bits then counting from left to right... me too...
look at the examples below and answer the multiple choice...

summarize 10.1.32.0 to 10.1.35.255?

        a)10.1.32.0/23
        b)10.1.32.0/22
        c)10.1.32.0/21
        d)10.1.32.0/20


summarize 172.168.12.0/24 to 172.168.13.0/24?

        a)172.168.12.0/23
        b)172.168.12.0/22
        c)172.168.12.0/21
        d)172.168.12.0/20

example 1
step 1: identify the octet where the action is.  the action is where the numbers change
in the first example the numbers change in the third octet, and like wise in the second example

.32 to .35 (you don't have to write down all the numbers, just the beginning and end of the range)
and .12 to .13

step 2: the first 2 octets are the same 10.1 and 172.168. together they comprise 16 bits so 16 is our working number

step 3: write down the first number .32 in example 1 and the last number where the change occurred, .35

.32
0010 0000

.35
0010 0011

count up to, but not including where the bits change and add 16 + 6  = /22
16 is the working number, determined by the address where there are no changes and 6 is the last position in the octet before a change

example 2
summarize 172.168.12.0/24 to 172.168.13.0/24?

the change happens in the third octet

     .12
0000 1100

     .13
0000 1101

16 plus 7 = /23
     
Class of address does not matter... the working number is always determined by the octet where the change occurs...

not sure about you, but there are a lot less 1's and zero's to count with my method... go ahead and prove this wrong...