Correct : a,b,c

Find the ISP''s Available Range
The ISP block is 202.16.0.0/15. The /15 means the first 15 bits are fixed.
- Let''s look at the 2nd octet (bits 9 through 16): The number 16 in binary is 00010000.
- Since 15 bits are fixed, the first 7 bits of this octet are locked as 0001000_.
- The 8th bit is free to be a 0 or 1.
- Therefore, the 2nd octet can range from 00010000 (16) to 00010001 (17).
- ISP Range: 202.16.0.0 to 202.17.255.255.
Determine the Required Client Block Size
The client needs 6,000 IP addresses. We need to find the number of host bits (n) where 2ⁿ ≥ 6000.
- 2¹² = 4096 (Too small)
- 2¹³ = 8192 (Sufficient)
The client needs 13 bits for the Host ID.
The Network Prefix is therefore 32 - 13 = /19.

Find the Valid Subnet Boundaries
For an IP address block to be valid, all of its Host ID bits must be zero.
- A /19 prefix means the Network ID uses the first 3 bits of the 3rd octet (8 + 8 + 3 = 19).
- The remaining 5 bits in the 3rd octet belong to the Host ID and must be set to 0.
- This means the value of the 3rd octet must be a multiple of 2⁵ (which is 32).
- Valid 3rd octet values: 0, 32, 64, 96, 128, etc.
Evaluate the Options
We must check if the 2nd octet is in range (16 or 17) AND if the 3rd octet is a multiple of 32:
- A) 202.16.0.0/19: 2nd octet (16) is in range. 3rd octet (0) is a multiple of 32. (Valid) ✓
- B) 202.17.64.0/19: 2nd octet (17) is in range. 3rd octet (64) is a multiple of 32. (Valid) ✓
- C) 202.16.32.0/19: 2nd octet (16) is in range. 3rd octet (32) is a multiple of 32. (Valid) ✓
- D) 202.17.24.0/19: 2nd octet (17) is in range. 3rd octet (24) is NOT a multiple of 32. (Invalid) ✗
Correct Answers: A, B, C