I have 6 tests under my belt and I saved 216 for the end. I have been\nstudying subnetting and thought I had it down. As I take practice tests it\nseems that is not the case.\n\nSo anyone that knows or can point me in the right direction/website/anything\nyour help would be much appreciated. I really want to know this not just\nfor the test but for my job in general.\n\nI will use my last practice question as an example of how I am confused.\n\nThe question wants to know which addresses can be divided into at least 4\nsubnets with at least 100 hosts per subnet.\n\nA. 10.25.0.0/22\nB. 10.25.0.0/23\nC. 10.25.0.0/24\nD. 10.25.0.0/25\n\n# My Calculations:\n\n/22 = 255.255.252.0 = 14 subnet bits 10 host bits\n/23 = 255.255.254.0 = 15 subnet bits 9 host bits\n/24 = 255.255.255.0 = 16 subnet bits 8 host bits\n/25 = 255.255.255.128 = 17 subnet bits 7 host bits\n\n\nusing 2^n -2\n\nA = 2^14 -2 = 16382 subnets and 2^10-2 =1022 hosts\nB = 2^15 -2 = 32766 subnets and 2^9-2= 510 hosts\nC = 2^16 -2 = 65534 subnets and 2^8-2= 254 hosts\nD = 2^17 -2 = 131070 subnets and 2^7-2= 128 hosts\n\nI thought the answer should be all of them. Apparently I was wrong.\nThe test says that only A and B are correct. It says that C can be divided\ninto 2 or 4 subnets and D can be divided into 4 subnets with only 30 hosts.\n\nI know it is a CIDR address but I guess I just don't quite get that concept\nor at least the math behind it. Is there a foormula for this type of\ncalculation? Am I just way off with subnetting and need to read the book\nagain? I don't quite understand where you start and stop counting the bits\nwith CIDR.