With a /27 subnet mask and subnet zero usable, which set of three IP addresses are all valid host addresses (not network or broadcast) in their respective /27 subnets?

Introduction / Context:
Subnetting questions often test your ability to recognize valid host addresses versus network and broadcast addresses. A /27 mask means each subnet contains 32 IP addresses, with the first address being the network and the last being the broadcast. This question asks you to identify a set of three addresses that are all valid host addresses within different /27 subnets.

Given Data / Assumptions:

• The subnet mask is /27, which corresponds to 255.255.255.224.
• Each /27 subnet contains 32 addresses (0 through 31 in the last five bits for the host part).
• Subnet zero is usable, so the first subnet block is not reserved.
• We must avoid any address where the host portion is all zeros (network) or all ones (broadcast) for its /27 block.

Concept / Approach:
For a /27 mask, the increment in the last octet is 32. Subnets begin at .0, .32, .64, .96, .128, .160, .192, and .224. In each block, the network address is the first (.0, .32, and so on) and the broadcast address is the last (.31, .63, .95, and so on). Any value between these two inclusive boundaries is a valid host address. To decide whether a given IP is a host address, compute the last octet modulo 32 and check whether it is neither 0 nor 31.

Step-by-Step Solution:
Step 1: For 10.15.32.17, the last octet is 17, which lies between 0 and 31 in the 32.x block and is neither 0 nor 31, so it is a valid host in the 10.15.32.0 /27 subnet.Step 2: For 66.55.128.1, the last octet is 1, which lies between 0 and 31 in the 128.x block and is neither 0 nor 31, so it is a valid host in the 66.55.128.0 /27 subnet.Step 3: For 135.1.64.34, compute 34 modulo 32, which is 2. The 64.x block spans 64 to 95; 34 falls in the next 32 wide block 32 to 63, but using the modulo method, 34 mod 32 = 2, so it is not at the boundaries 0 or 31, making it a valid host in the 135.1.64.32 /27 subnet.Step 4: For 17.15.66.128, the last octet 128 is exactly a multiple of 32, so it is a network address, not a host address.Step 5: For 129.33.192.192, the last octet 192 is also a multiple of 32 and is a network address for the 192.x block.Step 6: For 192.168.5.63, the last octet 63 is equal to 32 + 31, which is the broadcast address for the 192.168.5.32 /27 subnet, not a host address.

Verification / Alternative check:
Listing a few sample /27 ranges confirms this: for example, 192.168.5.32 to 192.168.5.63 has network 192.168.5.32 and broadcast 192.168.5.63, so anything between .33 and .62 is valid. Applying similar checks to each candidate address yields the same conclusion as the modulo method.

Why Other Options Are Wrong:
Option B includes 17.15.66.128 and 129.33.192.192, both of which are network addresses for their subnets.Option C consists entirely of non host addresses because 17.15.66.128 and 129.33.192.192 are network addresses and 192.168.5.63 is a broadcast address.Option D includes 192.168.5.63, which again is a broadcast address, making the set invalid.

Common Pitfalls:
Many candidates focus only on the last bits without computing where the subnet blocks start and end, leading them to mistakenly treat .128 or .192 as host addresses when those octets coincide with network boundaries. Always remember that in a /27, increments of 32 in the last octet mark network addresses, and the last address before the next increment is the broadcast.

Final Answer:
The only set in which all three addresses are valid hosts is 10.15.32.17, 66.55.128.1, and 135.1.64.34.