 Using your subnet worksheet, which octet value has an 8 on the left of the colon? Replace the leftmost zero in the native subnet mask with this number, and you have your answer: the subnet mask you need is 255.255.224.0. It's that simple. 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 net- work with 1,200 hosts in it? Here's the easy way to figure this one out. A /24 network has 256 hosts in it. Double this number until you get a number larger than 1,200: 256 X 2 = 512 512 x 2 = 1,024 1,024 X 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 36-16. The entries between the lines are subnet octet values that fall between classful boundaries. This third column of numbers in Figure 36-16 indicates a further binary progression. These numbers relate to the number of hosts in a subnet. You can keep the progression going as long as you need. When you get to the top of the sheet, start a new column to the left. These values will be the number of hosts for your second and first octets. IP Subnetting Made Easy 3rd Octet 4th Octet Value 1:256 2:128 16:16 Double this number 128:2 256:1 Figure 36-16. Subnet worksheet step #3 Looking at this worksheet reveals that the required subnet mask is 255.255.248.0. The network 172.16.0.0 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. Beware of a trick question you may see on tests. When using this method, you must remember that the first and last subnet may not be used given classical subnetting rules, and the first and last host may not be used. Therefore, if you encounter the question, "What would the subnet mask be for a network that required 1,024 hosts to be on the network?" the answer would not be 252, as it might appear, because two of those hosts are unusable (network and broadcast). If 