You manage 192.168.10.0 (a /24 network) and require at least nine subnets while keeping the maximum possible hosts per subnet. Which subnet mask should you choose?

Introduction:
Subnetting divides an address block into smaller networks. The design target here is to obtain at least nine subnets from a /24 while preserving as many host addresses per subnet as possible.

Given Data / Assumptions:

• Base: 192.168.10.0/24 (default mask 255.255.255.0).
• Required subnets: ≥ 9.
• Goal: maximize hosts per subnet subject to the subnet count constraint.

Concept / Approach:
Borrow the fewest bits that still meet the subnet-count requirement. From a /24, borrowing n bits yields 2^n subnets. For ≥9 subnets, n must be 4 because 2^3 = 8 (insufficient) and 2^4 = 16 (sufficient). A /24 + 4 bits = /28, whose mask is 255.255.255.240. Hosts per /28 = 2^(32-28) - 2 = 14 per subnet, which is the maximum host count achievable while still having at least nine subnets from the /24 block.

Step-by-Step Solution:
1) Determine bits needed: need ≥9 → n = 4.2) New prefix: /24 + 4 = /28.3) Mask: 255.255.255.240.4) Verify: 16 subnets, 14 hosts each.

Verification / Alternative check:
Trying /27 gives only 8 subnets; trying /29 gives 32 subnets but fewer hosts per subnet, violating the “maximize hosts” objective.

Why Other Options Are Wrong:

• /26 (255.255.255.192): 4 subnets (if using old rules) or 64 subnets in CIDR context across larger blocks, but within this /24 borrowing only 2 bits gives only 4 distinct /26s (insufficient for ≥9 from a single /24).
• /27 (255.255.255.224): 8 subnets, still short of 9.
• /29 and beyond: meet subnet count but reduce hosts further than necessary.

Common Pitfalls:
Mixing legacy “no subnet-zero” conventions with modern usage; today subnet-zero is allowed, but the math above already ensures ≥9 using /28.

Final Answer:
255.255.255.240