For the IPv4 host address 10.16.3.65 with a /23 prefix, what is the corresponding subnet (network) address?

Introduction / Context:
Subnetting is a key concept in Cisco networking and IP addressing. Knowing how to calculate the network address for a given host and prefix length is essential for designing and troubleshooting networks. This question asks for the subnet address for the host 10.16.3.65 with a /23 prefix, which means understanding how many bits are used for the network and which range of addresses fall into that subnet.

Given Data / Assumptions:

Host IP address is 10.16.3.65 in dotted decimal format.
The prefix length is /23, which corresponds to the mask 255.255.254.0.
We assume standard classless addressing without special reserved subnets beyond network and broadcast addresses.

Concept / Approach:
A /23 subnet mask means that the first 23 bits are used for the network portion and the remaining 9 bits are for host addresses. In dotted decimal, /23 is 255.255.254.0. The third octet is where the block size shows up. The block size in that octet is 256 minus 254, which equals 2. This means each subnet spans 2 values of the third octet. For a 10.x.x.x network, the subnets in the third octet will be 0 to 1, 2 to 3, 4 to 5, and so on. We need to find which block includes the third octet value 3, then identify the starting address of that block as the network address.

Step-by-Step Solution:
Step 1: Convert the /23 prefix length to a subnet mask. /23 equals 255.255.254.0.Step 2: Determine the block size in the third octet: block size = 256 - 254 = 2.Step 3: List the subnet ranges in the third octet: 0 to 1, 2 to 3, 4 to 5, and so on.Step 4: Since the third octet is 3, it belongs to the subnet whose third octet range is 2 to 3, so the network address is 10.16.2.0.Step 5: Attach the prefix length back to get 10.16.2.0/23 as the subnet.

Verification / Alternative check:
An alternative method is to write the third octet in binary. A /23 mask means the first 7 bits of the third octet are network bits and the last 1 bit of the third octet together with all 8 bits of the fourth octet are host bits. The third octet value 3 is 00000011 in binary. With a mask of 11111110 in the third octet, the last bit is a host bit and is set to zero in the network address. This forces the third octet of the network address to 00000010, which is 2 in decimal, confirming the network 10.16.2.0.

Why Other Options Are Wrong:
Option A, 10.16.3.0/23, is wrong because 3 is not the starting block value in a /23 network. The block including 3 starts at 2, not 3.
Option C, 10.16.2.0/24, is wrong because the prefix length changed to /24, which does not match the given /23 scenario.
Option D, 10.16.0.0/23, is wrong because the block 0 to 1 in the third octet would cover 10.16.0.x and 10.16.1.x, not an address with third octet 3.

Common Pitfalls:
A common mistake is to assume that the network address always ends with .0 in the last octet without considering the prefix length. Another pitfall is to think in terms of default classful boundaries and ignore classless subnetting, especially with prefixes like /23 that cut across octets. Students sometimes forget to calculate block size as 256 minus the mask value in the relevant octet, which is a quick way to determine subnet ranges.

Final Answer:
10.16.2.0/23