What part of 192.168.10.51 is the Network ID, assuming a default subnet mask?

Concept

An IPv4 address is split into a Network ID (which network the host belongs to) and a Host ID (which host within that network). With a default classful subnet mask, the class of the address — decided by the first octet — fixes how many leading octets form the Network ID:

• Class A (first octet 1–126): mask 255.0.0.0 → Network ID = first 1 octet.

• Class B (first octet 128–191): mask 255.255.0.0 → Network ID = first 2 octets.

• Class C (first octet 192–223): mask 255.255.255.0 → Network ID = first 3 octets, host = last octet.

Application

1. Read the first octet of 192.168.10.51 → it is 192, which lies in 192–223, so this is a Class C address.

2. The default mask for Class C is 255.255.255.0, so the masked (network) portion is the first three octets and the last octet is the host.

3. Apply the mask: keep the first three octets, zero the host octet → Network ID = 192.168.10 (full network address 192.168.10.0), and the host part is 51.

Cross-check

Reverse it: every address from 192.168.10.1 to 192.168.10.254 shares the same first three octets, so they must share one Network ID — 192.168.10 — confirming the host byte (51 here) is what varies. Hence the Network ID of 192.168.10.51 under the default mask is 192.168.10.