IPv6 practice help, confused

binaryhat

I'm doing Odom's practice examplee and I'm confused..

. 1)

3124::DEAD:CAFE:FF:FE00:1/80

3124:0:0:DEAD:CAFE::/80

why isn't it 3124::DEAD:CAFE::/80? 80/4=20 so the first 20 are reserved, the rest all 0's

. 2)

210F:A:B:C:CC CC:BOBO:9999:9009/40

210F:A::/40

40/4=10 so first 10 are reserved but I only count 8

3)

3124::DEAD:CAFE:FF:FE00:1/60

3124:0:0:DEA0::/60

60/4=15 so 15 are reserved I count 16 in 3124:0:0:DEA0::/60 and why DEA0? Should be DEAD?

4)

2BCD::FACE:1:BEFF:FEBE:CAFE/56

2BCD:0:0:FA00::/56

56/4=14 so 14 are reserved Again I count 16 in 2BCD:0:0:FA00::/56 and why FA00?

5)

3FED:F:EO:D000:FACE:BAFF:FE00:0/52

3FED:F:EO:D000::/52

52/4=13 are reserved I count 16 in 3FED:F:EO:D000::/52

Thanks in advance...

subnettingpractice

1) you cannot have two ::'s in an address. it creates ambiguity.
2) Two 0's are included in the end of the network bits. 210F:A:B:C:CC CC:BOBO:9999:9009/40 is 210F:000A:000B:C:CC CC:BOBO:9999:9009/40 with the bold part being the first 10.
3) You have to include the zero at the end: 3124:0:0:DEA0::/60. If you dont it reads 3124:0:0:DEA::, which is actually: 3124:0:0:0DEA:: the zero just shows where the DEA belongs.
4+5) same as 3.

Hope it helps.