Which subnet mask supports 500 hosts with fewest unused IP addresses?
Short answer
D 255.255.254.0
Question
Key Takeaways
This question tests your ability to calculate the smallest subnet that meets a host requirement without leaving excessive unused IPs, and the trap is confusing total addresses with usable addresses or choosing a larger subnet for expansion.
To support exactly 500 hosts while minimizing unused IP addresses, a /23 subnet mask (255.255.254.0) is required because it provides 512 total addresses and 510 usable addresses. Community experts agree that D is correct because any larger subnet would waste more addresses.
Common mistake: Choosing a larger subnet such as C (255.255.252.0 /22) because it comfortably supports 500 hosts, while forgetting that the question specifically asks to limit unused IP addresses; /22 wastes 522 usable addresses.
Expert Analysis
Why the Answer Is Correct
A /23 subnet has the mask 255.255.254.0, providing 9 host bits. With 9 host bits, 2^9 = 512 total addresses, of which 2 are reserved for network and broadcast, leaving 510 usable addresses - enough for 500 hosts with only 10 unused usable addresses. This is the smallest subnet that satisfies the 500-host requirement, so it best meets the goal of limiting unused IPs. Comments [1], [3], [5], and [6] all confirm that 255.255.254.0 /23 yields 512 total and 510 usable hosts, making it the correct match.
Why the Other Options Are Wrong
Option A (255.255.224.0 /19) provides 8,190 usable addresses, leaving thousands unused. Option B (255.255.248.0 /21) provides 2,046 usable addresses, wasting over 1,500 addresses compared to the requirement. Option C (255.255.252.0 /22) provides 1,022 usable addresses, still wasting 522 addresses; although it would support 500 hosts, it does not limit unused addresses as tightly as D. As comment [2] explains, the other subnets provide more addresses but result in a higher number of unused addresses, which violates the stated requirement.
Community Comment Notes
Multiple comments, including [1], [3], and [5], correctly identify D as the answer by calculating 9 host bits and 2^9=512 total IPs. Comment [3] makes an important point: the question does not mention expandability, so you should not assume a larger subnet is needed - the goal is the fewest unused addresses. Comment [2] explicitly compares the subnets and notes that options like /19 and /21 create many more unused addresses. The community is unanimous with 100% votes for D, reinforcing that the right method is exact-fit subnetting rather than over-provisioning.
Official Reference
- https://www.rfc-editor.org/rfc/rfc950
- https://www.rfc-editor.org/rfc/rfc4632
- https://www.iana.org/assignments/ipv4-address-space/ipv4-address-space.xhtml
These links open the official vendor documentation in a new tab.
Exam Strategy
Calculate the required host bits using 2^n - 2 >= required hosts. For 500 hosts, n must be at least 9, so the subnet prefix is /23 (32 - 9), which equates to 255.255.254.0. Avoid the trap of adding extra bits for future growth unless the question explicitly mentions expansion - always read for the 'limit unused addresses' requirement.
Community Votes
100% of anonymous learners picked answer D. Votes are pick records left by other test-takers — they are not the verified answer.
Discussion 5 contributions
Compiled from public community threads and reader submissions, lightly edited for clarity. Handles and vote tallies are shown as recorded, and comments reflect the views of individual contributors.
From public community threads 5
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a87d6a4 👍 3 Answered D255.255.254.0 (or /23). This subnet provides exactly 512 IP addresses, with 510 usable addresses for hosts. This minimizes the number of unused addresses while meeting the requirement to support 500 hosts. The other subnets, while providing more addresses, would result in a higher number of unused IP addresses: 255.255.224.0 (or /19) provides 8192 IP addresses, with many more unused. 255.255.248.0 (or /21) provides 2048 IP addresses. 255.255.252.0 (or /22) provides 1024 IP addresses. A /23 subnet is the most efficient choice for limiting unused IP addresses while meeting the requirement.
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mr_reyes 👍 3 Answered DTo get 500 hosts you need 9 host bits, thats gives us an mask of 255.255.254.0 /23. This gives us 510 hosts, with 508 useable addresses. They did not mention expandability, so we should not read that into the answer, they did however, asked to have the fewest unused addresses.
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Marinos_89 👍 3 Answered D/23 = 11111111 11111111 11111110 00000000 32-23= 9 2^9=512
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jcre 👍 2 Answered D255.255.254.0 = /23 provides 512 hosts (this includes the network and broadcast address) and 510 usable hosts.
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Marinos_89 👍 2 Answered D255.255.254.0 = /23 provides 512 hosts
Reader comments & corrections
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