I think that 192.168.0.0/22 is the proper solution ,is not it?

*1,453 posts since*

Sep 11, 2011

i have a different way to do it, that i think is easier… the first step is determining which octet the action is going to take place… group A is a bit of a distractor, however bear in mind that within group A the third octet is always 0 and in group B you have a range of 4 (but within a block of 8). since 0 and 4 is more than 4 numbers total it can’t be /22 because 2 to the second = 4 but there are 5 numbers to include, and five isn’t reachable with 2 to the 2nd… so to get 5 numbers would require 2 to the 3rd or 8… 2 to the 3rd will include 8 numbers total…

once you have determined the power of 2 that includes the numbers, just subtract the value of that power from the classful mask… /24 – 2 = /22 (remember the action is in the 3rd octet) but that only includes 4 numbers… /24 – 3 = /21 has to be correct because it takes 2 to the 3rd to include up to 8 numbers…