This is how you know i’m def gonna fail my university class this year :/
Because it is a very fun game and if it is fun people like it
Answer:
Check the explanation
Explanation:
223.1.17/24 indicates that out of 32-bits of IP address 24 bits have been assigned as subnet part and 8 bits for host id.
The binary representation of 223.1.17 is 11011111 00000001 00010001 00000000
Given that, subnet 1 has 63 interfaces. To represent 63 interfaces, we need 6 bits (64 = 26)
So its addresses can be from 223.1.17.0/26 to 223.1.17.62/26
Subnet 2 has 95 interfaces. 95 interfaces can be accommodated using 7 bits up to 127 host addresses can represented using 7 bits (127 = 27)
and hence, the addresses may be from 223.1.17.63/25 to 223.1.17.157/25
Subnet 3 has 16 interfaces. 4 bits are needed for 16 interfaces (16 = 24)
So the network addresses may range from 223.1.17.158/28 to 223.1.17.173/28
This is to much work for 5 points cmon
