shortening ipv6 question

[ibanez77]

Hello everyone
I've been studying for my CCNA and would like to know if i understand how to shorten the ipv6 correctly.

From what i understand the rules are these:
1: You cannot drop any trailing 0's in the address (2000 stays 2000)
2: You can drop any leading 0's in each individual blocks (0078 would be 78)
3: You can replace blocks of 0's with :0: (0000:0000:0000 would be 0:0:0)
4: You can drop continuous blocks of 0's with double colon :: but only once in the address. (2000:0000:0000:0000:0000:0100:0000
(this would then become 2000::100:0 )

ex:1: Given an address
2000:0000:0234:0078:0001:0000:0000:0001 (shortened to)
2000:0:234:78:1:0:0:1 or
2000:0:234:78:1::1 (this would be the shortest expression correct?)

ex:2:
3450:1234:0000:0000:0000:0101:0000:0000 (shortened to)
3450:1234:0:0:0:101:0:0 or
3450:1234::101:0:0 (this would be the shortest expression correct?)

ex:3:
FD00:0000:0000:0000:0000:0000:0100:0001 (shortened to)
FD00:0:0:0:0:0:100:1 or
FD00::100:1 (this would be the shortest expression correct?)

thanks for any and all help with this.:)