 This time we're going to be working with another device that also listens for the address BCDUDAD. But this time it ignores three different bits. It's going to be ignoring bits 9 through 11. As before, we saw that bits 0 through 3 corresponded to the first D. 4 through 7 corresponded to the A. And then 8 through 11 will correspond to the second D. So this time all of the bits that are getting changed are going to just be in that second D there. This time though, since we have three bits, we will have two to the third possible combinations or eight. So I'm going to begin by writing down eight different copies of BCDUDAD. But I'm going to leave out that middle D. So this will give me the ability to write down all eight possible combinations that this device can listen to. In this case, we know that we're not interested in lines 9 through 11, which correspond to the three most significant bits in this D. The rightmost bit in this digit is a 1 since it's a D. I'm going to go through and I'm going to set this digit to all of the possible variations where the last bit is a 1. That's pretty much going to be all of my odd numbers from 1 through 15. And somewhere around 13, I should have my D again and get BCDUDAD again. So 1, so I've got 0, 0, 0. 0, 0, 1, 1, I get 3. And I get 0, 1, 0, 1, which is 5, 7, 9. 9 plus 2 is B. So that's 1, 0, 1, 1. 1, 1, 0, 1 is D. And then 1, 1, 1, 1 is F. So in this case, this device listens to eight different addresses. And there is the list of all eight of those addresses.