An ISP having an address block 202.16.0.0/15 assigns a block of 6000 IP addresses to a client, using the classless internet domain routing (CIDR) super netting approach. Which of the following address blocks can be assigned by the ISP?

The correct options that can be assigned are 202.16.0.0/19, 202.16.32.0/19, and 202.17.64.0/19.

Here is the step-by-step logical reasoning and calculation to find the valid CIDR address blocks:

Step 1: Determine the block size required for 6000 IP addresses
In CIDR supernetting/subnetting, block sizes must always be powers of 2. We need to find the smallest power of 2 that can accommodate at least 6000 IP addresses.
- $2^{12}=4096$ (which is not enough)
- $2^{13}=8192$ (which is sufficient)

Therefore, the client must be allocated a block of 8192 IP addresses. This requires 13 bits for the host ID portion.

Step 2: Find the CIDR prefix length (subnet mask)
An IPv4 address consists of 32 bits. The subnet mask length is calculated by subtracting the host bits from the total bits:
$32-13=19$ bits.

Hence, any valid allocated block must have a mask of /19.

Step 3: Analyze the ISP's address block range
The ISP owns the address block 202.16.0.0/15.
- A prefix length of 15 means the first 15 bits of the address block are fixed.
- The first octet (8 bits) is fixed at 202.
- The second octet (next 7 bits) is fixed as 0001000 in binary. The 8th bit of the second octet is variable (either 0 or 1).
- This allows the second octet to range from 16 (binary 00010000) to 17 (binary 00010001).

Thus, the ISP can only allocate blocks whose second octet is either 16 or 17.

Step 4: Determine the valid starting addresses for a /19 block
For any /19 block:
- The first 16 bits cover the first two octets (which must begin with 202.16 or 202.17).
- The next 3 bits of the prefix belong to the third octet, leaving the remaining 5 bits of the third octet as host bits (which must be 0 for the network address).
- Since the 5 least significant bits of the third octet must be 0, the value of the third octet must be a multiple of:
$2^5=32$.

Therefore, the third octet of any valid /19 block must be one of the following multiples of 32:
0, 32, 64, 96, 128, 160, 192, 224.

Step 5: Verify the given options
- 202.16.0.0/19: Second octet is 16 (valid), third octet is 0 (valid multiple of 32). This is a valid block.
- 202.16.32.0/19: Second octet is 16 (valid), third octet is 32 (valid multiple of 32). This is a valid block.
- 202.17.64.0/19: Second octet is 17 (valid), third octet is 64 (valid multiple of 32). This is a valid block.
- 202.17.24.0/19: Second octet is 17 (valid), but the third octet is 24, which is not a multiple of 32. This is invalid.