 Alright, so let's take a look at another way we can use the Euclidean Algorithm. And this is in a particular problem. And again, a useful thing to keep in mind is that a good way to fail a math course is to try and work problems by following the examples that somebody gives you. You don't learn math by watching somebody else do math. The only time you really should look for an example for insight in how to solve a math problem is after you've spent some amount of time trying to solve the problem on your own. And the reason that this is important is that the time you spent trying to solve a problem gives you an idea of what you've learned. It reviews things that you should know and causes you to pull together all the knowledge that you've acquired thus far. And this is an important part of the learning process. But after you've given it a good try to attempt to figure out a problem, sometimes you just need that extra little bit of help. And so, well, let's take a look at that. Let's take a look at this problem. We have m equal to something 6k plus 8 and n equal to 3k plus 1. And we want to assume that m and n are not relatively prime. And we want to find the greatest common divisor. And so, before proceeding, it's worth trying to figure out what we can't actually do. And so, let's take a look at that. One of the things we can't do is because we don't know what n and m are, because we don't have them as a product. We have them as a sum. We have them as a sum. We can't actually factor m and n. We can't determine what they are products of. And so, this means that finding the greatest common divisor is going to require us to use the Euclidean algorithm. Now, again, another way to fail mathematics is to use formulas and algorithms without understanding why they work. So, we know how to use the Euclidean algorithm, but it's important to remember why the Euclidean algorithm works. And the reason that it works is that anything that divides both numbers will also divide the remainder when I divide one by the other. That means I'm going to want to take a look at the remainder when we divide m by n. Well, again, another way to fail mathematics is to use formulas and algorithms without understanding why they work. There is a division algorithm. There is a way that we can divide two numbers, but how do you apply it to something like this? Well, again, we want to go back to why that algorithm works. And in particular, because I need the remainder, well, it's the amount that I have left after I've removed the divisor as many times as possible after I've subtracted it. So, I can do division by repeated subtraction at 6k plus 8. I'll remove the divisor once to get that. And I can remove it again. And I have my remainder. And at this point, I don't have anything left over, so now I can try and find the greatest common divisor. It's got to be something that divides the remainder. So, there's my remainder of 6. And so, I know that my greatest common divisor has to be a divisor of 6. And I can list my choices. That's 1, 2, 3, or 6. Now, we know that the numbers aren't relatively prime. Because you know the definitions, you can do mathematics. They're not relatively prime, which means that 1 is not the greatest common divisor. Not relatively prime means they do have a greatest common divisor larger than 1. So, I can eliminate 1 as a possibility. Now, the next step takes a little bit more observation and insight, but we might notice here that n is a 3k plus 1 number. Well, what does that mean? Well, by the definition of division, I know that n divided by 3 is k with remainder 1. That's our definition of division with quotient and remainder. So, I know that n divided by 3 is k with remainder 1, which means that 3 is not a divisor of n. And in particular, if 3 isn't a divisor of n, there's no possibility it can be a common divisor. So, I know that 3 is not the greatest common divisor. So, I can get rid of that. And in addition, if 3 is not the greatest common divisor, I can also get rid of any multiple of 3, which means that 6 also is not a possibility. And so, our greatest common divisor, one of the divisors of 6, well, the only one that works in the circumstance is going to be 2. And so, the greatest common divisor must be 2.