N10-005 · Question #578
Which of the following uses eight groups of characters for addressing?
The correct answer is C. IPv6. IPv6 uses 128-bit addresses represented as eight groups of four hexadecimal digits separated by colons.
Question
Which of the following uses eight groups of characters for addressing?
Options
- AMAC
- BIPv4
- CIPv6
- DDHCP
How the community answered
(19 responses)- C95% (18)
- D5% (1)
Why each option
IPv6 uses 128-bit addresses represented as eight groups of four hexadecimal digits separated by colons.
MAC addresses use 48 bits expressed as six groups of two hexadecimal octets, not eight groups.
IPv4 addresses use 32 bits expressed as four dotted-decimal octets, not eight groups.
IPv6 addresses are 128 bits long and written as eight groups of four hexadecimal characters (e.g., 2001:0db8:85a3:0000:0000:8a2e:0370:7334), satisfying the requirement of eight groups.
DHCP is a protocol for dynamically assigning IP addresses and does not define an address format with eight groups.
Concept tested: IPv6 address structure and format
Source: https://learn.microsoft.com/en-us/previous-versions/windows/it-pro/windows-server-2008-R2-and-2008/dd379516(v=ws.10)
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