Hello, Can someone please help me with the following. I understand everything up to the part about "the remaining 5 bits identify the hosts on that subnet". Here is the question and answer: Your company has been assigned a network ID of 197.132.240.0. Your network has 5 subnets. The max number of hosts per subnet is 25. Which subnet mask should you use for the network? A. 255.255.255.192 B. 255.255.255.224 C. 255.355.255.240 D. 255.255.255.248 Correct answer is B. A network ID of 197.132.240.0 is a class C network. A class C network falls in the range of 192.0.0.0 to 223.255.255.0. The high-order bits of a class C network ID are always set to 1 1 0. The next 21 bits (completing the first three octets) identify the remaining portion of a class C network ID. In this case, the first three octets of the network ID are 197.132.240, which translates to a binary value of 11000101 10000100 11110000. The default subnet mask of a class C network is 255.255.255.0 (binary 11111111 11111111 11111111 00000000), indicating that the first three octets define the network ID. The remaining octect can be used to define which bits should be used to identify the network's subnetting structure and which should be used to identify the host on those subnet. For a class C network, a subnet mask of 255.255.255.224 (binary 11111111 11111111 11111111 11100000) indicates that the first three octets and the first 3 bits of the last octet identify the subnetted network ID. The remining 5 bits identify the hosts on that subnet. Because 3 bits are used to define the subnetting structure of this class C network, the network will support up to 8 subnets and up to 30 hosts per subnet. I understand everything up to this: Because 3 bits are used to define the subnetting structure of this class C network, the network will suport up to 8 subnets and up to 30 hosts per subnet. Can someone please elaborate on this for me? This is what I understand about calcing subnets so far: Binary is a Base2 system VS. Decimal is a Base10 system. This means that all binary numbers can be expressed as 2n. Which means each 8-bit field matches this chart: 128 64 32 16 8 4 2 1 27 26 25 24 23 22 21 20 Use this chart to convert number between decimal and binary. Example: What is the binary value of 153? 128 64 32 16 8 4 2 1 27 26 25 24 23 22 21 20 1 0 0 1 1 0 0 1 Or 128+16+8+1=153 Classes: Class A addresses have the highest level bit (27) turned off. So the address must be less than 128. Class A range = 0-127 Class B addresses have the highest level bit (27) turned on and the 2nd highest-level bit (26) turned off. So the addresses will differ from Class A by the first 64 bytes. Class B range = 128-191 (127+64=191) Class C addresses have the highest level bit (27) and the 2nd highest-level bit (26) turned on and the 3rd highest- level bit (25) turned off. So the addresses will differ from Class B by the first 32 bytes. Class C range = 192-223 (191+32=223) # Of networks and hosts: Points to remember - The 0. and 127. networks are used for loopback and must be subtracted from the Class A networks. - Each 8-bit field has a maximum value of 255, but to avoid the possibility of all 1's or all 0's subtract 2 when computing the number of networks and hosts. Class A Net: 0-127= 128-2= 126 networks Hosts: 253x253x253=16 million Class B Net: 128-191=64 (different first bytes) x253=16,000 networks Hosts: 253x253=64,000 hosts Class C Net: 32 (different first bytes) x253x253=2 million Hosts: 253 hosts It's this last part to calc the hosts that's baffling me. Any help to understand this would be greatly appreciated. Thanks in advance, BB