Subnetting: correct address for router interface?

Question

This is a portion of a larger network, but I've simplified it for the sake of the question.

Here is the network.

Adresses in it are as follows:

Server: 192.168.0.97/30
Router to server IF0: 192.168.0.98/30
Network with 8 pcs: 172.16.40.144/28
The 8th pc: 172.16.40.152/28
Router to the network with 8 pcs IF1: 172.16.40.158/28

Now, I've been told that last one is incorrect, but I can't figure out why. As far as I understand it, it should be the last available adress of the .144 network, thus .158

The teacher figured out it was wrong looking at this table.

But looking at it now I can't find his reason.

Any help appreciated.

Answer 1

The last 4 bits of the 0th pc on the network would be 0000, which converts to 172.168.40.144. The last 4 bits of the highest possible value is 1111, which converts to 172.168.40.159. I'm not sure if that would be the broadcast address though.