Solution

The correct answer is 4

Key Points

• Based on the previous question where we established the new subnet mask as /16 resulting in 64 addresses in each subnet:
• The first address of each subnet is a multiple of 64 in the last octet, and the last address is 63 greater than the first address (because we start at 0).
• The last subnet can be calculated by subtracting 64 (one subnet's worth of addresses) from the entire address space. So the last subnet's first address can be calculated as 256 - 64 = 192 (the "256" being representative of the total number of addresses accessible by 8 bits in the last section of the IP address), with the "192" being the starting point of the final subnet in the /26 partitioning.
• Hence, the first address of the last subnet is: 130.56.255.192
• The last address in the subnet is the first address plus 63, because there are 64 addresses in each subnet. That would be: 130.56.255.255

So, the answer is: 4) First address = 130.56.255.192, Last address = 130.56.255.255.