DRAG DROP
You are configuring a multi-subnet IPv6 network for a regional office.
The corporate network administrator allocates the 2001:0db8:1234:0800: :/54 address space for
your use.
You need to identify network IDs of the first and last subnets that you will be able to create at the
office.
Which network IDs should you identify?
To answer, drag the appropriate network IDs to the correct subnets. Each network ID may be used
once, more than once, or not at all. You may need to drag the split bar between panes or scroll to
view content.
Answer: 
10.10.1.0/26
10.10.16.0/25
10.10.128.0/23
10.10.8.0/22
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Dear god, I think I’ve finally got it.
You are given a /54 address to work with.
Normally with IPv6 (eight blocks in total), the first 48 bits (first three blocks) are used for global networking, the last 64 bits (last four blocks) are used for your hosts and thus you have 16 bits left (in the fourth block!) to do your subnetting with.
However your enterprise admin has ‘eaten’ away 6 of these 16 bits to do some subnetting already.
This leaves you with 10 bits left.
Now lets look at that fourth block in the given address. It’s: 0800 in hex.
1) convert this to decimal
0000 1000 0000 0000
2) count from the left how many of those bits are already dedicated to the network (6 in this case) and keep those separate.
0000 10 00 0000 0000
3) the part on the right is thus yours to play with. Convert al 0’s to 1’s to get the last possible subnet address, first in binary.
0000 10 11 1111 1111
stuck back together it looks like:
0000 1011 1111 1111
4) now convert this back to hexadecimal! Joy!
bin: 0000 1011 1111 1111
dec: 0 11 15 15
hex: 0 B F F
So the last possible fourth block looks like: 0bff.
And the answer is /64 because of a standard rule in using IPv6 subnetting.
Re-read page 210 in the 70-410 book for the explanation.
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Uh, step 1) should of course say ‘convert this to binary’, my bad.
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