Find the network: What is the subnetwork number for host 172.16.66.0/21?

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:

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