Subnet reasoning — interpret 10.16.3.65/23 and choose the correct statements Given host IP 10.16.3.65/23, evaluate the following statements about its subnet address, lowest/last host, and broadcast address, and select the two true statements.

Introduction / Context:Understanding CIDR masks lets you compute subnet ranges quickly. With /23, each subnet spans 512 addresses (510 usable). Determining network, broadcast, and host ranges is fundamental for addressing, ACLs, and summarization.

Given Data / Assumptions:

• IP: 10.16.3.65/23.
• Mask: 255.255.254.0 (since /23 = 8+8+7 bits).
• We consider standard usable host rules (exclude network and broadcast).

Concept / Approach:With /23, the third octet increments in steps of 2. Subnet ranges are 10.16.0.0–10.16.1.255, 10.16.2.0–10.16.3.255, 10.16.4.0–10.16.5.255, etc. Address 10.16.3.65 falls in the 10.16.2.0/23 subnet. Therefore, the subnet (network) address is 10.16.2.0, the broadcast is 10.16.3.255, the first usable host is 10.16.2.1, and the last usable host is 10.16.3.254.

Step-by-Step Solution:Compute block size: 256 – 254 = 2 in the third octet.Locate range that contains third octet 3 → the 2–3 block.Network address: 10.16.2.0; Broadcast: 10.16.3.255.Usable hosts: 10.16.2.1 through 10.16.3.254.Validate statements: (1) claims subnet 10.16.3.0 → false; (2) lowest host 10.16.2.1 → true; (3) last host 10.16.2.254 → false; (4) broadcast 10.16.3.255 → true.

Verification / Alternative check:Convert the /23 mask to binary and AND with the IP to confirm network bits. A quick sanity check: broadcast equals next subnet start minus one (10.16.4.0 − 1 = 10.16.3.255), which matches.

Why Other Options Are Wrong:

• A/C/D/E: Each includes at least one false statement (wrong network or wrong last host).

Common Pitfalls:Mixing /23 with /24 rules and assuming third-octet boundaries always align on 1-octet blocks. Always compute the block size for the variable octet.

Final Answer:2 and 4