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Given the MAC Address 4F:9C:56:2B:40:E3 and an IPv6 prefix of 2001:DB8:1:2, what would the IPv6 address generated using EUI-64 look like?
2001:DB8:FFCE:1:2:4F9C:562B:40E3/64
2001:DB8:1:2:4D9C:56FF:FE2B:40E3/64
2001:DB8:1:2:4F9C:562B:40E3:FFFE/64
4F9C:56FF:FE2B:40E3/64
2001:DB8:1:2:40E3:56FF:FE2B:4F96/64
EUI-64 inserts FF: FE into the middle of a 48-bit MAC address and inverts the seventh bit of the first byte value of the MAC address (which is 'F') to make it 64-bits long. Adding a 64-bit network prefix to the beginning gives you a full 128-bit IPv6 address.
Add FFEE to the middle of the MAC Address which turns into 4F:9C:56:FF:FE:2B:40: E3
Flip the 7th bit--4F equals 01001111 and becomes 01001101 which equals 4D
Host ID is now 4D:9C:56:FF:FE:2B:40:E3
Full IPv6 EUI-64 ID is 2001:DB8:1:2:4B9C:56FF:FE2B:40E3/64
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