Stumped!

Found this from one of the forums that another poster posted. There seems to be an obvious error on the question since nothing I do can match the explanation of the correct answer at the bottom:

11.) Which of the following IP addresses fall into the CIDR block of 110.68.4.0/18? (Choose three.)

A. 110.68.8.32
B. 110.68.7.64
C. 110.67.6.255
D. 110.66.3.254
E. 110.65.5.128
F. 110.64.12.128

Answer: B, C, E. A Class A network address with a /18 is 255.255.192.0. The subnets in the third octet are 0, 64, 128, 192. The network address in the question is 110.64.0.0, with a broadcast of110.64.127.255, since the next subnet is 110.64.128.0. Answers B, C, and E are correct host IDs.

Here's how I solved it:

Prefix /18 --> 11111111.11111111.11000000
Decimal equivalent: 255.255.192.0

The zero subnet for the given network should be 110.68.0.0. Based on what we know of Network Classes, this is a class A network with the first octets defining the network. In the explanation given, it mentions that the network address is 110.64.0.0. That doesn't make sense to me because even though it is a Class A network, the first two octets can't change since the first two octets of the given mask is 255.255.

Using Wendell Odom's technique in the ICND1 exam guide:
256 --> Total number of decimal values in an octet
-192 --> decimal value of the masks 3rd octet.
= 64 --> The subnet blocks for the 3rd octet.

Therefore the range of IP addresses for this subnet block are:
110.68.0.0 --> Subnet number
110.68.64.0 --> Next non-zero subnet

Therefore the valid IPs for host are 110.68.0.1 - 110.68.63.254. The only IP addresses in the choices given that fall under this range are choices A and B.

Am I wrong on this?!?!

update (03.11.08):
Link to a few responses from Techexams.net forums:
http://www.techexams.net/forums/viewtopic.php?t=32073