 Okay, let's consider divisibility from another viewpoint, and in this case we're going to rely, again, on the distributive property, and again, this is just the divisibility of linear combinations. And it's based off of the distributive property, so suppose I have d dividing two other numbers, then for any integers a and b, then d is also going to divide what's called the linear combination, a times the one of them, plus b times the other one. And the proof of this is relatively straightforward by definition of divisibility of d divides m, then m is d times x, and likewise, n is d times y. So consider that linear combination, a m plus b n. So a m, a times dx, b n, b times dy, add them together, and I get this, and my distributive property says that this common factor of d can be removed, and this linear combination is d times something, which says that d is a divisor of that linear combination. And this theorem is the basis for a bunch of divisibility rules that help us determine easily whether or not a given number is divisible by certain other numbers. Now, in order to make use of these rules, in order to express these rules, it's convenient to write a three-digit base 10 number as a, b, c, where this time we're going to read each of these values here as the digits of the number. So this is actually going to be the number a hundred b tens c ones. So it's useful to ask ourselves the question, well, what does that really mean? Well, a number like a, b, c, that's a hundred b tens c ones, that's a hundred times a, plus 10 times b, plus c. And the one thing not to get confused by is that here a, b, c is indicating that a, b, and c are the second digits of our three-digit number. Meanwhile, 100a refers to the multiplication 100 times a, 10 times b, and then c by itself. The important thing here is over on this expression, we can do arithmetic as normally. This expression is a three-digit number, and we can't actually do the arithmetic that we've been working with using this expression. It won't make any sense. So let's consider divisibility by 2 and by 5. The first thing to notice here is this three-digit number a, b, c. I can rewrite that 5 times 20a, 5 times 2b, plus c, and I note that 5 divides the first term because it's 5 times something, also the second term, again, 5 times something, which tells me that if 5 divides the last term, 5 will also divide this quantity. And if I put that observation together, this gives me my divisibility rule for 5. If 5 divides the last digit of a base 10 number, then 5 will divide the number. And analogously, if I rewrite this three-digit number 100a plus 10b plus c as 2 times something plus 2 times something plus c, then 2 divides this, 2 divides this. If 2 divides the last digit c here, then 2 is going to divide the number, again, only in base 10, or at least the reason that we can make this statement goes back to the idea that our three-digit number a, b, c can be written this way and can be broken apart into something that is divisible by 5 or is divisible by 2. So these divisibility rules always are base-related. Now, just as a note, if I have a number with 4 or more digits, it's the same argument that 4-digit number is going to be 1,000 times something plus 100 times something plus 10 times something and so on. It splits into things that are divisible by 5 or things that are divisible by 2. And so, what can I do? Well, I'll consider a number n and I'll just write down some random number. And if I want to determine whether it's divisible by 2 or by 5, I look at the last digit. And that last digit is 5 is divisible by 5, so the number is divisible by 5. Likewise, the last digit is not divisible by 2, so the number itself is not divisible by 2. And in the application of the divisibility rules, there's not a whole lot of interesting problems that arise from it. We pretty much get answers of yes or no. The more interesting question is, how do we get the divisibility rules in the first place? So again, let's consider, again, a 3-digit number. This time, we're going to consider divisibility by 3 and by 9. And this time, we'll do a little bit of algebra. This 3-digit number, A, B, C, 100A, 10Bs, and C, I'm going to rewrite it this way. That's 99A plus an A, 9B plus a B, and then my C is left over, and I'll rearrange it. 99A plus 9B plus A plus B plus C. And the thing to notice here is that this first set of terms, I could rewrite 3 times something, 3 times something. So this set of terms here, divisible by 3, divisible by 3, and then whatever's left over. And so we can make the same argument as before, because 3 divides this. If 3 divides this, then 3 is going to divide our original quantity, which is our 3-digit number, A, B, C. And without going through the details, because they should be fairly obvious, 9 divides this, 9 divides this. So if 9 divides this, we'll also divide our number. So we now have a divisibility rule for 3 and for 9. If 3 divides the sum of the digits, then 3 divides the number. Likewise, if 9 divides the sum of the digits, 9 divides the number. And again, this is only true if we're working in base 10. So for example, suppose I wanted to determine whether 3 or 9 divides this number. So we'll take a look at the sum of the digits, plus 8, plus 9, plus 1, and I add those together, I get 21, which is divisible by 3, so 3 divides this amount. Here's an interesting question you might want to consider. How can you tell whether 21 is divisible by 3? Well, you could actually do the division, but what worked once works again. If you wanted to determine whether this is divisible by 3, add the digits, and see if you get something that's divisible by 3. So 2 plus 1 is 3, and we're all set. So 21, however we note, is not divisible by 9, so 9 does not divide 3,891. So what about divisibility by 6? Well, since 6 is 2 times 3, our factor theorem tells us that if a number is divisible by both 2 and by 3, then it's going to be divisible by 6. So here's a number, and let's see if it's divisible by 6. So we'll check. The last digit is 2, so it's divisible by 2. The sum of the digits, 3 plus 7 plus 2 plus 4 plus 2 plus 6 is 24, and if you don't know what's divisible by 3, 2 plus 4 is 6, and that is divisible by 3. So this is divisible by 3, and so is n, and so now n is divisible by 2 and by 3, so it's also going to be divisible by 6. For divisibility by 4 and 8, we might note that our 3 digit number I can express is 100a plus 10b plus c, and 4 divides this first term under day. So if 4 divides the rest of these things, then 4 is going to divide our 3 digit number. The thing to notice here is this last bit here, this is b, 10s, and c. This is really the number that we would form by taking the last 2 digits of our 3 digit number. So this gives us the divisibility theorem if 4 divides the number 4 by the last 2 digits of the number, then 4 will divide the number. For 8, it's convenient to think about 4 digit numbers abcd, that's 1000a plus 100 plus 10 plus d, and this first term is 8 times 125a. So 8 divides the first term, so if 8 divides the rest, 8 divides the number, and again, this is the number b100c10d, it's the number formed from the last 3 digits of our number. So that gives us our theorem for divisibility by 8 if 8 divides the number formed from the last 3 digits, then 8 divides the number. And again, the examples aren't too exciting, so here's the number, and we want to determine whether 4 and whether 8 divides it, so we look at the last 2 digits, that's 24, and 4 divides 24, so 4 divides n. Likewise for 8, this last 3 digits form the number 224, and 8 divides 224, so 8 divides n once again. The only one we're missing is divisibility by 7, so we have divisibility theorems for division by 2, 5, 3, and 9, 6 we can handle by determining whether something's divisible by 2 and also by 3, and we have divisibility by 4, divisibility by 8, so the only one we're missing at this point is divisibility by 7. Unfortunately, in base 10, there's no easy way of determining whether a number is divisible by 7, so the question of whether a number is divisible by 7, it's probably just easier to actually do the division and see if it comes out evenly.