Okay I just went through this chapter at school this past week so I am really new at this myself. But after going over your post and working my way through it I think I might have an answer.
In your example, 126.96.36.199 --> MAC = 01:00:5E:59:E6:45
there is only one possible IP address besides the one above that will match the MAC address. As you stated you take the last 23 bits of the IP address and convert it into the last 24 bits of the MAC address. The first bit from the second octet (128) is always translated as 0 into the MAC address.
Thus, the only two IP addresses that would translate to that MAC are 231.217 and 231.89 (which is 217-128).
11011001 = 217
01011001 = 89
Since the first bit is always translated to the MAC as 0 both of those numbers are converted to 59 in hexidecimal form.
Now, since the range of multicast addresses are from 188.8.131.52 - 184.108.40.206 you would have two addresses in each network that would translate to 01:00:5E:59:E6:45
They would be 220.127.116.11 and 18.104.22.168, followed by 22.214.171.124 and 126.96.36.199, and so on up until 188.8.131.52 and 184.108.40.206.
The only problem with that is it only gives you 30 possible IP addresses that translate to the same MAC address. Where do the remaining two come from?
The administrator has disabled public write access.