I know this is kind of old and you probably already turned in the assignment but in case you don't actually understand the concept:
Basically you want to maximum amount of users for the 7 divisions. so a /9 (255.128.0.0) is only enough for 2 divisions. And a /24 (255.255.255.0) can accompany 7 divisions but you have very few users and have a lot of unused addresses. So the goal is to find that perfect ground where the subnet mask(s) can encompass as much of your given block of 10.0.0.0/8 as possible while having a minimum requirement of 7 divisions (networks)
Your given network mask looks like this (Class A) in binary:
11111111.00000000.00000000.00000000 (255.0.0.0)
then you just keep adding ones from left to right until your network can have at least 7 subnets:
Example:
/9 (255.128.0.0) (11111111.10000000.00000000.00000000)
Can have 2 subnets of:
10.0.0.0/9 and 10.0.0.128/9
So the equation is just 2^n where n = subnet bits (do not include Class network bits in this). This is is the equation to determine how many different networks you have for a given subnet mask. you need 7 . so 2^2 = 4 and 2^3= 8.
So the 3rd bit is the minimum needed to allow for 7 networks which is a subnet mask of 255.224.0.0
those 8 subnets would be (2nd octet only): 00000000 (0) , 00100000 (32), 01000000 (64), 01100000 (96), 10000000 (128), 10100000 (160). 11000000 (192), 11100000 (224).
Note: I am assuming that the assignment called for as many addresses possible in each subnet. Otherwise there are many answers.
This post was edited by kasey21 on Sep 6 2018 09:24pm