Which aggregate of the IPv6 addresses 2001:0303:0000:5000:0000:052B:0000:0000/96 and 2001:0303:0000:5000:0000:052C:0000:0000/96 has the longest possible mask?
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A. B. C. D.C.
To determine the aggregate address with the longest possible mask, we need to find the common bits in the two IPv6 addresses.
The two IPv6 addresses are:
To find the common bits in these two addresses, we need to convert them to binary form. Each block of four hexadecimal digits in an IPv6 address represents 16 bits in binary form. So, we can convert these two addresses to binary form as follows:
Now we can compare the two addresses bit by bit from left to right to find the common bits. The first 96 bits are the same in both addresses, so the aggregate address would be:
However, this question is asking for the longest possible mask, so we need to find the next smaller aggregate address with a longer mask. To do this, we need to look at the bits to the right of the first 96 bits in the two addresses. The first bit to the right of the 96th bit is different in the two addresses (i.e., 0 in the first address and 1 in the second address). So, we need to use a mask that includes the first 96 bits and the first different bit to the right, which is the 97th bit. This means we need a mask of /97.
To find the aggregate address with a mask of /97, we need to zero out the bits to the right of the 97th bit in the two addresses. This gives us:
This is the smallest possible aggregate address with a mask of /97 that includes both of the original addresses. However, this is not one of the options given in the answers. The closest option is C, which has a mask of /93. To check if this is a valid aggregate address, we need to convert it to binary form and check if it includes both of the original addresses: