what would be the last legitimate host address in this …

You have been assigned a network ID of 172.16.0.0/26. If you utilize the first network resulting from this ID, what would be the last legitimate host address in this
subnet?

A.
172.16.0.64

B.
172.16.0.63

C.
172.16.0.62

D.
172.16.0.65

Explanation:
When a class B address such as 172.16.0.0 is subnetted with a /26 mask, the subnet mask in dotted decimal format is 255.255.255.192. This means that the
interval between the network IDs of the resulting subnets is 64. The resulting network IDs are as follows:
172.16.0.0
172.16.0.64
172.16.0.128
172.16.0.192
172.16.1.0
and so on.
For the network ID 172.16.0.0, the last address in the range is 172.16.0.63, which is the broadcast address. Neither the network ID nor the broadcast address for
any subnet can be assigned to computers. This means that the addresses that can actually be assigned range from 172.16.0.1 to 172.16.0.62. The last legitimate
host address, therefore, is 172.16.0.62.
172.16.0.63 cannot be used because it is the broadcast address for the 172.16.0.0 network.
172.16.0.64 is the network ID for the 172.16.0.64 network, and 172.16.0.65 is the first address in the second network.
Objective:
Network Fundamentals

Sub-Objective:
Apply troubleshooting methodologies to resolve problems

Cisco > Support > IP Routing > Design TechNotes > Document ID: 13788 > IP Addressing and Subnetting for New Users