From: John Mistichelli (jmistichelli@xxxxxxxxx)
Date: Wed Jul 17 2002 - 13:26:57 GMT-3
Sure,
For the first two addresses only the first octet
needed wildcard bits. Convert to binary:
140 = 10001100
161 = 10100001
Assuming a number line of 128,64,32,16,8,4,2,1 the
bits that are not in common between the two are 32, 8,
4 and 1. If you just add those together you get 45.
All other octets had to match exactly. Hence 45.0.0.0.
I used the same method on the other two only setting
wilcard bits in the second octet.
59 = 00111011
63 = 00111111
All bits are common except the 4 position. Hence
0.4.0.0.
Hope that helps.
John
7536
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