Find the network: What is the subnetwork number for host 172.16.66.0/21?
Networking
Subnetting
Difficulty: Easy
Choose an option
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A172.16.36.0
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B172.16.48.0
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C172.16.64.0
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D172.16.0.0
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E172.16.72.0
Answer
Correct Answer: 172.16.64.0
Explanation
Introduction / Context:Summarization-style masks like /21 require spotting the third-octet block size and then choosing the correct boundary that includes the given host.
Given Data / Assumptions:
- IP: 172.16.66.0/21
- /21 mask = 255.255.248.0
- We need the corresponding network address.
Concept / Approach:For /21, the block size in the third octet is 256 - 248 = 8. Valid third-octet boundaries are 0, 8, 16, ..., 64, 72, etc. We select the boundary less than or equal to the host’s third octet (66).
Step-by-Step Solution:
Third-octet block size = 8.Find boundary ≤ 66: ... 56, 64, 72 ... → 64 is the correct lower boundary.Thus, the network is 172.16.64.0; the broadcast is 172.16.71.255.Verification / Alternative check:Range check: 64 ≤ 66 ≤ 71 confirms 66.0 lies within the 64.0/21 block.
Why Other Options Are Wrong:
- 36.0 and 48.0: Different /21 blocks; do not contain 66.0.
- 0.0: Too broad; not the /21 block for this host.
- 72.0: Next block above; does not include 66.0.
Common Pitfalls:Forgetting that /21 spans multiple Class B /24s; miscounting the 8-value steps in the third octet.
Final Answer:172.16.64.0