Given the IPv4 address 192.168.10.51 and the default Class C subnet mask (255.255.255.0), which portion represents the Network ID?

Introduction / Context:
IPv4 addressing separates an address into a network portion and a host portion using a subnet mask. With default classes, a Class C network uses a /24 mask. This question checks identification of the Network ID from a typical private address.

Given Data / Assumptions:

• Address: 192.168.10.51.
• Default Class C mask: 255.255.255.0 (also written /24).
• Network ID is formed by masking the host bits to zero.

Concept / Approach:

Applying a /24 mask means the first three octets identify the network, and the last octet identifies the host. Therefore, the Network ID corresponds to 192.168.10.0. Some representations omit the trailing .0 when asking for the network portion, effectively referring to the first three octets as the Network ID component.

Step-by-Step Solution:

Verification / Alternative check:

CIDR notation 192.168.10.0/24 confirms that any host 192.168.10.x shares the same Network ID.

Why Other Options Are Wrong:

192: Only the first octet; insufficient.

0.0.0.5: Not meaningful for network identification.

51: This is the host identifier within the /24 network.

None of the above: Incorrect because 192.168.10 is the correct network portion among choices.

Common Pitfalls:

Expecting the answer to include .0 explicitly. Conceptually, the network is 192.168.10.0/24; the option captures the network portion.

Final Answer:

192.168.10