The subnet mask for a particular network is 255.255.31.0. Which of the…

Key idea: compute the network address by bitwise AND between each IP and the subnet mask. Apply the mask octet-by-octet.

Note: The mask 255.255.31.0 is not a valid contiguous-ones subnet mask because the third octet (31) in binary is 00011111 (ones are not contiguous). Nevertheless, if you apply the given mask literally (bitwise AND), you can determine which addresses share the same masked network address.

Mask (per octet) in binary: 255 = 11111111, 255 = 11111111, 31 = 00011111, 0 = 00000000.

• 172.57.88.62 and 172.56.87.233 — The mask forces the first two octets to match (255). The second octets are 57 vs 56, so the network addresses differ; not the same network.

• 10.35.28.2 and 10.35.29.4 — First two octets match (10.35). For the third octet: 28 & 31 = 28, and 29 & 31 = 29, so the masked third octets differ; not the same network.

• 191.203.31.87 and 191.234.31.88 — Third octet masked gives 31 for both, but the second octet is part of the network (mask = 255) and 203 ≠ 234, so these are in different networks.

• 128.8.129.43 and 128.8.161.55 — First two octets match (128.8). Third octet: 129 & 31 = 1, and 161 & 31 = 1, so the masked network address for both is 128.8.1.x (fourth octet masked by 0). Therefore these two addresses are on the same network under the literal mask.

Final answer: 128.8.129.43 and 128.8.161.55 are in the same network when 255.255.31.0 is applied literally. However, because 255.255.31.0 is not a valid contiguous subnet mask, clarify the intention with the question author; if a standard /21 mask (255.255.248.0) was intended, these addresses would not be in the same subnet.