damewoodr,
Glad you're enjoying ITProTV and you're asking a great question!
The answer lies in your subnet mask! If you were to write out your subnetmask of the /22 = 255.255.252.0 This last octet is our only real concern here.
If you were to begin with a 10.0.0.0/8
One network 10.0.0.0, 16,777,214 hosts
With your subnet mask of /22, this means that you can have 2^14 number of subnets= 16,384 possible subnets from that 10.0.0.0/8 network. (I will not do all of them but I will set the pattern of where this begins.
1st - 10.0.0.0/22
00001010.00000000.00000000.00000000/22
2nd - 10.0.4.0/22
00001010.00000000.00000100.00000000/22
3rd - 10.0.8.0/22
00001010.00000000.00001000.00000000/22
4th - 10.0.0.0/22
00001010.00000000.00001100.00000000/22
5th - 10.0.4.0/22
00001010.00000000.00010000.00000000/22
6th - 10.0.8.0/22
00001010.00000000.00010100.00000000/22
...
(notice the interval of 4 in between the subnets--this comes from the fact that we are not concerned with 10 hosts bits) but we are concered with the FIRST 22 bits.)
If you were to copy and paste the above binary into a word proccessor and continue to work it all the way through to 10.55.40.0/22...
Xst - 10.0.40.0/22
00001010.00000000.00101000.00000000/22
Xnd - 10.0.44.0/22
00001010.00000000.00101100.00000000/22
Xrd - 10.0.48.0/22 00001010.00000000.00110000.00000000/22
Xth - 10.0.52.0/22 00001010.00000000.00110100.00000000/22
The /22 give you an interval of 4 because the 22nd bit is in the 3rd octet with a place value of 4. So your network ID numbers will always be increments of 4.
Hope this explaination helped you.
Cordially,
Ronnie Wong
Host, ITProTV