Consider three machines M, N, and P with IP addresses 100.10.5.2, 100.10.5.5, and 100.10.5.6 respectively.

Solution:

First, we will do bitwise AND between Subnet mask and given IP address one by one

For M: 100.10.5.2

Bitwise AND:

    01100100.00001010.00000101.00000010 (100.10.5.2)

AND 11111111.11111111.11111111.11111100 (255.255.255.252)

___________________________________________________________

   = 01100100.00001010.00000101.00000000 (100.10.5.0)

For N: 100.10.5.5

Bitwise AND:

    01100100.00001010.00000101.00000101 (100.10.5.5)

AND 11111111.11111111.11111111.11111100 (255.255.255.252)

___________________________________________________________

   = 01100100.00001010.00000101.00000100 (100.10.5.4)

For P: 100.10.5.2

Bitwise AND:

01100100.00001010.00000101.00000110 (100.10.5.6)

AND 11111111.11111111.11111111.11111100 (255.255.255.252)

___________________________________________________________

   = 01100100.00001010.00000101.00000100 (100.10.5.4)

It is sure from the above computation that N and P belong to the same network (i.e., 100.10.5.4), while M can belong to the same network or it can be from any other network.