 So how does the division method work? Well, as the complement to the multiplication method, you can kind of expect that it's going to be pulling out factors and looking at those. So while I'm starting with the number in decimal, I'm going to consider what it would actually look like if I'd represented it in the destination base. So, if I've got, say, A times 5 to the fourth plus B times 5 cubed plus C times 5 squared plus D times 5 to the first plus E. So, perhaps I have this number compacted into decimal form. And I'm going to divide by my destination base, which in this case is 5. So, when I divide by 5, each of these coefficients is something less than 5. I want to find out what they are. And dividing by 5 is going to pull a 5 out of each of these terms. All of these are bigger than 5 because they've got an exponent of 5 as part of them. The E though doesn't have any exponents of 5. It's something less than 5. So, when I do this division, I'll get A times 5 cubed plus B times 5 squared plus C times 5 to the first plus D. Now, D is out of 5. We factored out all of the 5s from D times 5 to the first. And the E didn't have any factors of 5 in it at all. So, I get a remainder of E. Now I have the least significant digit from my result. And I can go back and I can start again and do division on my new quotient. So, I'll divide this by 5. So, A times 5 cubed divided by 5 gives me A times 5 squared. I have this term, B times 5 squared divided by 5 will give me B times 5 to the first. C times 5 to the first divided by 5 is just C. And then again, D is something less than 5, so I can't divide it by 5. There won't be anything left afterwards. So, now I'm just going to get a remainder of D. And I have a new quotient. I can work with this again. So, I'll get A times 5 to the first plus B with a remainder of C. 5 this by 5. I'll get A with remainder of B. And now I'll get 0 for remainder of A. So again, I can read this down from top to bottom. And I'll get A, B, C, D, E. But what I've done here is I've pulled out factors of 5. And anytime I discover that I can't pull out a factor of 5, that must be the coefficient of the next exponent. First I found coefficients for 5 to the 0, then 5 to the first, 5 squared, 5 cubed, lastly 5 to the fourth. So, this method works by finding those factors kind of iteratively. Each time we pull out a factor of our base, we're left with some remainder. That remainder tells us the coefficient of the next term in our list. And we just keep working down our list until we run out of things that can have factors of our base. We're left with a 0 and we're done.