 So one of the important things we might look for is known as the GCT. And this emerges as follows. Given two or more numbers, a common divisor is a divisor that is common to all of the numbers. And it divides all of the numbers, in other words. The greatest common divisor is the largest of these common divisors. And so what we call the GCD, the greatest common divisor, we tend to start by looking among the common divisors. And then we can pick out what the greatest of those common divisors is. So for example, let's look at the GCD of n equals this and n equals this number. And a little analysis of the problem goes a very long way. Because I want to find the greatest of the common divisors of m and n, because I want to find the largest number that is a divisor of m and n, I might want to start by looking for any divisor of m. So let's see what we can find. Well, because we know our definitions, we can do the mathematics. A divisor d is a divisor of m if and only if m is the product of d and something else. So I know what m is as a product, 2 to the 3rd, 3 to the 2nd, 7 to the squared. Well, I can write that out, m looks like this. And because of associativity and commutativity, we can group these factors any way that we want to. We can write this product of 2s, 3s, and 7s in any order of the 2s, 3s, and 7s that we want. The only thing we can't do is we can't drop any out and we can't write in new ones. So, well, suppose I take these factors, this 2, this 2, this 3, and this 7. So I'll take those factors. And m is going to be the product of those factors and the remaining factors, the 2, the 3, and the 7. And so m is the product this times this. And so that tells me that this 2 times 2 times 3 times 7 is a divisor of m. And likewise, 2 times 3 times 7 is also a divisor of m. Well, I could take this set of factors, this 2 and this 7, and I'll use those. And then I'll take the remaining factors, the 2, 2, 3, 3, 7, and I'll do those because I have all the factors I had originally. This is also going to be equal to m, except this time I can see that 2 times 7 is a divisor of m. And likewise, I can take these factors 2, 7, and 7, and that's going to be one set of factors. And then I can take the remaining factors. And so that gives me the remaining factors and I can put those together. And again, that gives me m as a product and 2 times 7 times 7 is going to be a divisor. Well, what that means, a little analysis goes a long way. I can find a divisor of m by taking some, all, or none of these prime factors. And likewise, I can do the same thing for n. I can form its divisors by taking some, all, or none of the prime factors of n and multiplying them together. So here's an important idea. It doesn't matter that our first guess is not the correct guess as long as we eventually get to what we want. And so I want to find the greatest common divisor. So I might start by looking for any divisor. And one of the things I know is that one divides any number. So one is a common divisor of m and n. It's not the greatest, it's not necessarily the greatest common divisor, but it's a good starting point. All right, so let's see. Well, we do see that 2 appears in both prime factorizations. So 2 is going to be a divisor of m and also a divisor of n. So I can include that in my greatest common divisor. Actually, we can go a little bit farther. m has this factor 2 to the third. That's 2 times 2 times 2. n has this factor 2 to the second, 2 times 2. So both of these have a factor of 2 times 2. So my divisor of m and n, I can include 1, 2, but I can actually include 2, 2s, because both of them have at least 2, 2s. Well, let's take a look at the next one. 3 appears in both prime factorizations. So I can include 3 as part of my common divisor and the same argument as before. m has 2 factors of 3 and n has 3 factors of 3. So both of them have a factor of 3 times 3. So I can include another 3 in there as well. Well, since 7 does not appear in both prime factorizations, it can't be part of the common divisor. It divides m, but it does not divide n. It is not part of the common divisor. And there's no other prime that appears in both. 5 doesn't appear in m, so 5 can't be a common divisor either. And since we've included everything that could be a divisor of both and nothing that couldn't be, what we have here is the greatest number that's going to divide both. It's going to be 1 times 2 times 2 times 3 times 3. An important idea to recognize is that all of this relies on having the prime factorization of the numbers because that's the only time we can guarantee when a number does or does not divide one of two numbers. Because 7 is in this prime factorization, I can guarantee 7 divides m. Because 7 is not in this prime factorization, I can guarantee 7 does not divide n. I can't say that the same thing is going to be true if I have a non-prime factorization. So if I have p equals this, q equals this, well, we might try to do the same thing. We can find the common divisor. 1 divides both numbers because 1 divides every number, so we'll include 1, and 25 isn't prime, so we can't use the fundamental theorem of arithmetic, which means we should find the prime factorization of both numbers. So let's see, p equals 25 times 14. 25 factors is 5 times 5. 14 factors as 7 times 2. And so there's my prime factorization of p. Let's write that in a slightly more useful form. That's 2 times 5 to the second times 7. q, 35 times 28. Well, 35 factors as 5 times 7. 28 factors as 2 times 2 times 7. And again, I'll write that in exponential form to compact that a little bit. So there's my prime factorization of p and q. And so now I can start to look for the things that are common to both of them. So p has a factor of 2, and so does q. And there's only one factor of 2 that I can use. Both of them have 5 as a factor. This one has a single 5. This one has two 5s. So 5 by itself is the largest power of 5 that'll divide both. And 7 appears in both ones here twice there. So I can get 1 7 in the GCD, but I can't get any more. And so there's my greatest common divisor, 1 by 2 by 5 by 7. What happens if we have the greatest common divisor of 1,392 and 1,407? Well, just like the previous one, we have to find the prime factorization of p and q. Except that's very, very hard. Factoring is extremely time-consuming. It is by far the worst way of finding the greatest common divisor. This factorization only works if we have numbers that are ridiculously easy to factor. 25 is easy to factor. 14 is easy to factor. 35 and 28. These numbers are ridiculously easy to factor. So factorization works. But factorization is extremely time-consuming, and so we might ask, is there a better way of doing this? Yes, and we'll take a look at that next time.