er: the subnet mask you need is 255.255.224.0. It's as simple as that. Now, let's try a more complex problem, like one you might see on an exam. With a network of 172.16.0.0 255.255.0.0, what subnet mask would you need to allow a network with 1,200 hosts in it? Here's the easy way to figure this one out. A class C network has 256 hosts in it. Double this number until you get a number larger than 1,200: 256 * 2 = 512 512 * 2 = 1,024 1024 * 2 = 2,048 Let's fill in the sheet some more. Draw a line under the top entry, and another above the bottom entry, as shown in Figure 34-16. The entries between the lines are subnet octet values that fall between classful boundaries. This third column of numbers in Figure 34-16 indicates a further binary progression. 3rd Octet 4th Octet Value 1:256 2:128 16:16 Double this number 128:2 256:1 Figure 34-16. Subnet worksheet step #3 IP Subnetting Made Easy These numbers relate to the number of hosts in a subnet. You can keep the progres- sion going as long as you need. When you get to the top of the sheet, start a new column to the left. These will be the number of hosts for your second and first octets. Looking at this worksheet reveals that the required subnet mask is 255.255.248.0. The network 172.16.0.0 (a native class B network) with a subnet mask of 255.255. 248.0 will result in 32 subnets with 2,048 hosts in each. To validate the math, perform this equation: 8 * 256 = 2,048. A quick word about a trick question you may see on tests is in order. When using this method, you must remember that the first and last subnet may not be used, and the first and last host may not be used. Therefore, if the question, "What would the subnet mask be for a net- work that required 1,024 hosts to be on the network?" were posed, the answer would not be 252, as it might appear, because two of those hosts are unusable (network and broadcast). If you require 1,024 hosts to be live devices on the network, you must have a network capable of supporting 2,048 hosts. This is one of the limitations of usi