 So, the subtraction method allows me to do my arithmetic in the source base. So, it's really good for converting from, say, decimal into binary or octal or hexadecimal, but I won't really want to use it when going the other direction. So, for example, 25 in base 10, if I want to convert this to something in binary, I'm going to be looking for exponents of 2. So, I know there aren't any 32s in 25, but the next thing down is a 16, so I can pull out a 16. Tract 16, that leaves me with 9. The next exponent down of 16 would then be 8, and 8 is less than 9. So, I can pull out an 8. This leaves me with 1, which means I'm not going to have any 4s. I'm not going to have any 2s, but I will have a 1. So, I get 1, 1001 for 25, which is what we see over here. If I want to convert that to octal, I should expect to get 31. Hexadecimal, I'll expect to get 19. So, this time, if I start with 25, and I'm looking for powers of 8, I got 1, 8, and 64. So, I know I don't have any 64s in here, but I do have some 8s. How many 8s? It looks like I can get three 8s out of here. Three times 8 is 24. So, I'll subtract that, and I'll get 1. So, the first digit in my octal number will be 3. Then, I've got 1 left over, so 1 times 8 to the 0th would give me 1, so I'll put a 1 in here. And I get 31 for my octal number, which is, again, what we expected. For hexadecimal, I'm going to expect to get 19 out, because that's what's on my number line. But again, I'm starting with 25, and I'm going to try to take powers of 16 out. So, my first few powers of 16 are going to be 1, 16, and 256. I'm sure there are no 256s in my little 25, but I can't find a 16. Only 1, though. So, I'll subtract that 16. That will give me a 1 here. Give me a remainder of 9. Then, my next exponent of 16 is 1, so I can do 9 times 1. And that will be 9. So, I'd subtract 9, and I'd have 0 left over. The result here is that I get 19 in base 16, which is what I expect to get for 25 in decimal. If we pick a different number, like, say, 29, then we'll do the same sort of thing. If I'm converting this number to binary, I'm going to look for powers of 2. So, I know 32 is larger than 29. The next one down is 16 again. So, I'll subtract that. That will leave me with 13. And I will put a 1 in my 16 place. Now, the next exponent down from 16 is 8. So, I'll subtract 8. This will leave me with 5. I'll put a 1 in the 8 position. Now, I know I can pull a 4 into 1 out of here. So, I'll put a 1 in the 4th position, a 0 in the 2's position, and a 1 in the 1's position. This is the binary number that I get for 29, which matches up with what we've got here. Converting this to octal, it's the same thing. 29. Again, I've got 1, 8, and 64 as my smallest exponents here. I can't pull any 64s out, but I'll be able to pull some 8s out. 3 times 8 is 24, but 4 times 8 is 32, so I'll have to live with 3 times 8. So, I'll write down a 3 for my 8th position. 29 minus 24 leaves me with 5. So, I'm left with 5 times 1. I'll write down 5. And I get 29 is 35 in octal. And last, if I convert between 29 and base 10 and hexadecimal, I'm pulling out exponents of 16. So, I'll have 1, 16, and 256. I can't pull any 256s out of here, and I can pull 116 out. So, I'll subtract 16, write down a 1 for my 16th position. So, leave me with 13. And 13 is less than 16, so it is a valid number in hexadecimal. So, I just have to convert this 13 in decimal into hexadecimal. And what number do I get? I get D. So, my hexadecimal number then is 1D.