Yeah, so I got all answers E, but we have more than that. So, yeah, the answer is one of these. Yeah, E is a bit too small, so let's see if you could adjust your solution. So obviously, if n is always a square, right? I mean, a few things that we want to remember is, for example, I have an a to the 10th. We can rewrite s is a to the fifth square, right? So that's one thing. And then a to the, for example, to the square or to the fourth, a to square. And then it doesn't matter. I can raise this to the maybe the sixth or the seventh power. Yeah, I can raise it to an odd power. So that thing, which is a to the 14th. And then I can also rewrite this as a square, a to the 7th and also a square, right? So there are two ways to get a square. n to the nth is a square. The first case is that if n is, obviously, n is even, right? So if n is even, n equal to 2m, then n to the n, which is equal to, I can put 2m to the 2m. And then this one, I can rewrite it as 2m and then to the m and the whole thing squared, right? So for even n, then n to the n is definitely a square, right? So that would be the first observation. And we have 50 of them from 2, 4, also way to 100. So we have 50 even numbers. And after that, what do we have? What about odd numbers? Even odd numbers, n to the n, can it be a square? For example, if we consider 3, 3 square, so that would be 3 square. And then if we raise it to the 9th, then it really doesn't matter because I have a square over here, then I can rearrange it to 3, 9, to square, right? And that means for n odd. But if it is square numbers, then I would also have a perfect square, n square, 2 square, 3 square, all the way to 3, 9 square is 21. So I have another 5 odd numbers here. And the answer is B, 55 numbers, such that n to the n is a square of a natural number. So that's what we'll be doing today is a warm-up question. Our topic today is the number theory. So let's see what we have. This is our question, lesson number 3. So number theory, as you know, studies integer and their properties, right, like odd and even, if it's a perfect square, perfect cube, how far the prime numbers are apart from each other, consecutive prime numbers apart from each other, things like that, right? So the popular topics that we've seen in number theory in these levels are prime factorization, multiples, divisors, greatest common divisors, least common multiples, a lot of different divisibility rules, algebra with integers. That's something with our special focus for today. And things that you've seen earlier, like place value, decimals, and fractions can also come up again at this level. So let's go. The first one, prime numbers and prime factorization. So this one, everyone knows it by heart, right? Every positive integer has exactly one prime factorization. And some other important facts, like two integers whose only common positive divisors, one are relatively prime, for example, like 15. I have a 15 and maybe 64. These are, most of these are composite numbers, but because there should be one, so they are relative prime, and we use this a lot in problem solving. And then how to check if a prime is a number is a prime. So we have a theorem that we can actually prove that in order to check a number n is a prime or not, we only have to check whether n has any prime divisor less than or equal to square root of n, and we do not have to go all the way to n. So for example, we are given a 1,001, and we're asked whether they're prime. So this is where sort of arithmetic comes into play. We sort of remember that 1, 0, 2, 4 is a 2 to the 10, which is 2 to the fifth square, which is 32 square. So we know that we only have to check for prime factors up to 31, right, like all the way to 29. But we do not have to check for more than that. And actually, this number is also a good number to remember. 1,001 is actually a product of 7, 11, and 13. So when it comes to prime, to number theory, it's so important to remember. Like, I would remember all the prime numbers up to 100, all the prime factorization, like numbers that has a lot of divisors, like 60, 70, 2, 96. All of it, you want to remember by heart. You want to remember squares up to 20 square, the first 10 powers of 2, first few powers of 3 and 5, and some big prime numbers. And just do a lot of prime factorizations that really sort of, when you see the problems, that would help you get started right away, rather than spending time doing, remembering all the factors and factorizations. So the next thing is LCM and GCD. This is a common formula. But it's pretty obvious. After you factor A and B, then we just look at the common, the same. The GCD of A and B would have the same prime factors. And the powers would be the smallest of these two powers. And while the LCM would get the bigger of the two powers. So that would be obvious. Once you can factorize them, then you can find LCM and GCD pretty straightforwardly. And then divisibility of rules. This is also most of the rules that you already know. So odd, even, divisible by 3 is when some of the digits is also a multiple of 3. For example, over here, 5 and 7 is 12, 6 and 9. So this number is a multiple of 3. Divisible by 4, when the last two digits is multiple of 4, because this is going to be multiplied by 100. If it's divisible by 6, then you want to check whether it's multiple of 2 and 3. For 8, we want to check the last three digits, because this part is going to be multiplied by 1,000. For 9, the sum of the digits must also be a multiple of 9. So here is 9, this 9, so this number is multiple of 9. And then for 11, this is actually a very good and useful rule that got used a lot. You want to sum up all the numbers in the odd positions and then subtract the sum of the numbers in an even position, so it's like alternative position. So over here, we have 4 and 9. And 4, that's 17. And subtract away 1 and 5, that gives us 11, which is a perfect power, multiple of 11. So this number is multiple of 11, just like 1, 0, 0, 1. So the difference is 0. That's why it is multiple of 11. So let's first do a few questions on this, and we move on to the next topic. So first question, an arbitrary two-digit number consists of the digits a and b. By repeating this pair of digits three times, one obtains a six-digit number. This new number is always divisible by which of these options? So we would have a, b. That's what it means by two-digit numbers. And we repeat it three times. It's going to be a, b and a, b. So I'll give you a couple of minutes. Then you can put the answer into the chat. I'm sorry, into the poll. Or have any questions, ask the poll, and Andrew and I in the chat. okay so be careful with 11 right because 11 if you sum up going to be a and b and b so that means for 11 you would have to the sum in the odd and the even the difference be 3b minus 3a right so um when you say that we have to prove that this is always divisible by 11 that means this must always be divisible by 11 this is a shorthand notation for you know divisible and for all values this is also a shorthand notation for all values of a and b but is it true can we find example of a and b in in which this difference is not a multiple of 11 right so for example b equal to 1 and a equal to 0 is a counterexample where this difference is not a multiple of 11 so when we say that it's always divisible by one of these numbers so that means it must be true for all values of a and b and for example you want to show that it's not divisible by 2 then you should have to give one example and that's enough to cross out that choice okay i hope the reasoning is clear so that means 11 we can already cross out because one zero one zero one zero is actually not a multiple of 11 so that one is out okay yeah i give it a little bit more time You can just go through each choice as one strategy, like eliminate the obvious one and then narrow down choices. So among these, we have a divisible rule for, okay, great, I started to get correct answers. So let's do the easy one. 11 is a hard one, but we're about two. Two, we just need to get b odd, right? So just like one, three, one, three, one, three. So that one also out because it's an odd. So notice that we actually have the rules for most of these numbers. Five also easy, but as long as b is not zero or five, then it's not divisible by five. So that also out. And then for nine, now I can reach up because it's gonna be three times a plus b, okay? So in order for it to be a multiple of nines and a plus b must be a multiple of three, right? So you want to find a counterexample. You take a equal to one, b equal to four, for example. So one, four, one, four, one, four. You sum up, it's not multiple of nines. So that means it's out and 11 is out. So this is an example in which you can just eliminate all the incorrect choices, right? Elimination. And then, you know, the answer is c, okay? That's one way to get the answer in the exam. And another way to get it that it has to do with, you know, the topic of today later on, we call it, it's called integers of algebra, algebra of integers. So for this one, we use algebra and we actually can expand it this way. It's going to be a, b times, this is going to be, you know, 1,000, right? 10, is it a, b, okay, four, zero, and then plus a hundred. This, the first a, b give you 10,000. The second a, b gives us a hundred, right? And the last a, b give us one. So this sum is going to be one, zero, one, zero. Zero, one. And this one is actually a multiple of seven. We divide by seven, it's going to have a one. 31 by seven, we have four. 30 by seven, another four. 21 by seven, that is three, okay? So this one is actually seven times one, one, one, four, four, seven. This is an example in which you can use algebra together with number theory to get the answer. And that would be the perfect, like sort of rigorous proof, but most of the time you can get the answer. You know, in a different way. So I hope it's clear for everyone. Okay, if you have any question, just feel free to stop me and ask in the chat, or, you know, you can always stay at home. I have to do it in an hour, but I have time after that. So if you have any question, you can, you're welcome to stay and Andrew and I will help. Okay, Andrew, could you please do this question? So if the ratio between two positive integers, a and b is equal to three to two, and the ratio between the least common multiple a and b and the greatest common divisor a and b is equal to which of the following? If some students already got the correct answer, maybe you can give a hint and wait a little bit more for the rest to give it a try. So because we know the ratio is 3 to 2, we know that we can express A and B as 3 of some number and 2 as some number. And this should help us find the other two things we need. But yes, it's more like a generalized form, right? We also use algebra in this case. And then if you need another hint, you could also do, you know. Okay, so if you remember our definitions for greatest common number and least common multiple, we can use the definitions to find it in terms of the number. So because we know that least common multiple will take the highest exponent of each prime factor, we know that because 3a is three times more than a and 2a is two times more than a, we know that the coefficient 3 is going to be greater than a and the coefficient of 2 is going to be greater than b, so we know that the least common multiple is going to be 6a. And for greatest common denominator, we can see that 3 and 2 are co-prime, so we can't get that out. So we can take the least of both, which is going to be a. So now we have to find the ratio between 6a and a, which is 6. Okay, great. Thank you, Andrews. That would be like the generalized form is the one that captures all the information given to us, right? So that would be sort of like the ideal solution, but if you have difficulty finding out the solution in the exam, you know that the answer should not depend on the actual values of a and b, but only depends on the ratio. So you can always choose a specific example and check, right, like 3 and 2, the LCM would be 6 and the GCD 1, right? So the ratio is actually 6. Or you can do 6 and 4, and LCM is 12 and the GCD is 2, also 6. So really you sort of can move on with the exam and chose the correct answer. It's not the rigorous proof, but you know, you have to have the answer and maybe probably can inspire you to go and get this general solution. Okay, so algebra in interiors, like, you know, we've actually seen both of these two questions, we have, each question has two different ways to do that, and the general ways always use algebra. So just like what we did in last class, every time we have a problem, we have unknowns, we can assign variables to unknown and write a set of equations, and the same techniques applies to question in number theory. For example, if we have to find a four-digit number that has certain properties, then we just denote it ABCD, and then write a set of equations for the digits that satisfy the given properties, right? So, for example, if we say it's a multiple of 9, then you have one equation where the sum is a multiple of 9, and then it is multiple of 11, then we know that A plus C minus B plus D is a multiple of 11, and plus with, you know, other properties, other restriction, and then also the digit would be in the range from 0 to 9, then most of the time we can figure out the answers. And then we also have a set of equations called Diophantine equations, and basically it's just a fancy name for equation where the coefficients are integers and, you know, the solution, and we are only interested in part, in integer solutions, right? So, for that, for these equations, divisibility rules relatively prime inequalities, like when we have constraints and we don't forget to use inequalities, because a lot of time we have fewer equations than the number of variables, but because we have restrictions, we can use inequalities and all of these tools to help find the answers. And then some very common useful things in algebra in general, like manipulation of algebra expressions and algebraic factorization are also very helpful in solving, you know, Diophantine equations. For example, you know, this is a difference of square is the product of the difference and the sum, or factorize, things like that, super helpful in solving these kind of equations. And then earlier you probably have seen a lot of problem with the numbers of digits and unit digits, then, you know, divisibility rule and finding a patterns are some of useful tools that we can use to solve these kind of questions. So, let's go with more examples. So, this one. Each of the numbers a and b is a square of an integer. The difference a minus b is a prime number. So, which of the following could be b, right? So, again, you see a and b's are integer, but more than that, they are square of an integer. And the difference is the prime number, right? So, there are lots of unknown here, but, you know, in your mind, you probably can assume that we might be able to write a set of equations using some forms of a and b, a generalized form of a and b. That's one way to go forward. So, machine time 26. Sometime I got a question from one student and I will just send the answer to the whole class. Yeah, but of course you don't know who asked the question so, you know, feel free to ask questions. Oh great, I think we're making a lot of progress here. I just started with giving some hint and, you know, you have time to continue working on it. So again, we can use a generalized form. It's going to be x squared when x is an integer, b is y squared, also an integer, right? And then so we have, when you write it this way, then right away we see that there's a lot of restriction between these two because the difference of two squares must be a prime, right? p, we usually denote p as a prime, and not all of them have that. And I don't know how some of you got the answer, but one way is to do that. So for example, you can go through each choice. That's one of the, if it helps, I'm not sure how, but that's one way you can, you know, go through each choice. Then y is going to be equal to 10. So what is x, what is p, something like that, right? One answer I got is b, b is not the correct answer, so if it's your answer you may want to adjust it. So I give you one more hint, x squared minus y squared to b squared, and remember we can factorize this one, right, x squared minus square, the rule is identity is x minus y times x plus y, right, this is when identity comes into play, and I mean without that I think it would be hard to solve this question. So a prime number is a product of two integers, so that gives, that puts a lot of restrictions on those two numbers, right, what's the restriction x minus y and x plus y such that p is a prime number. Okay, so p is a prime number, this is smaller, this is bigger, so the only way for it to be prime is that this one has to be equal to one, right, and this one equal to p, so that means x and y, a and b are actually two consecutive prime numbers, right, so there would be a lot of, you know, x and y that would, maybe not a lot, but quite a number of them that would make this product a prime number, but if we go to the choice over here for a, we have y equal to 10 and x going to be 11, right, because x is one more than y, but x plus y is, okay, so I put here would be y and x and x plus y, so for choice a is going to be 10 and 11, and that's 21, but 21 is a composite, so a is out, right, and then we do 144, so it's 12, and this is 13, and the sum is 25, which is also a composite, so that's out, 256, so we remember that is 16, right, 16 squared, so 17, and this is 32, it's a multiple of 3, so that's out, 900 is 30, and 31, and now we are in good luck, right, because 30 plus 31 is 61, that is a prime number, and then for 100, I mean, we can stop here, but that one would be 100 and 101, and 201 is divisible by 3, right, because the sum of the digit is 3, so that one also out, so I don't know, maybe some of you would remember, I mean, one way you can do that, for example, you go through this, and then you check 30 is, maybe some students, they just have a good guess or good number skills, and you realize that, you know, 31 squared somehow become 961, and then you subtract this, and you got 61, and here's the answer, right, you know, could be, you could be solve it this way, and that means have a good number of skills, but I think that this one would be more straightforward, so this is an example when we use algebra in number theory. Yes, any questions, feel free to do in the chat, and I will answer it, look at it. Okay, question number four, the positive integer n is such that the product of its digit is 20, which of the following could not be the product of the digits of n plus one, right, so this is a question with the products of the digits is 20, and as in could not be, so okay, so it's similar to one, to the previous one, showing that it could be, you can only have to show one example, right, it's how the question is phrased, and could not be, then you need to show a general argument, and a lot of time it's harder to make a general argument, it's easier to to show one particular example, so, you know, keep that in mind when you do these questions. Here are just a few hints here. You can factorize 20. Definitely construct some examples. And yeah, I got some answers. Not all are correct. And then just make sure that you choose one, but you quickly go through all the others and make sure they are not the correct answer. It's good to double check. And in this case, actually, you could show one example. It's actually easier to show an example for the choices that are possible. Actually, you can construct a number n such that the product of the digits of n plus 1 is one of those choices. And for our five choices. And in the process of doing so, you would also find a general argument why the fifth one is not the correct one. And I would say it's easier to go with construction, you see, sort of construct the answer. So I give a little bit more time for the rest of the class to have a chance. So, for example, I do 20, right, it would be, the answer would be 4 times 5, right, and then I would have number n and n plus 1. If n is 45, then n plus 1, 46, and 4 times 6 is 24, right? So I have 24, that's choice E, so actually E is possible. And then I do 54, we can just switch the order, so that would be 55, and then I have 25, right, so this is sort of like easiest option, 24 and 25, that I could do. And then you move on, the answer would be harder, but, you know, we could also do 225, for example, 225, and then, for example, this is 225, but that's no good, 225, 226, 227, 226 is still 24, we can, you know, stir it up a little bit, 5, 2, 2, right, and that would be 5, 2, 3, so if you add this, it's going to be 5, 2, 3, so this is 10, and 3 is 30, so actually, you know, the question looks scary, because, you know, how many numbers that, it's talking about n and n plus 1, right, and these are not products, these are addition, things like that, but actually, when we look at it closely, we say there's actually not many choices of n whose product of the digit is 20, and that would be 30 is also possible, when we are left with 40 and 35, and, you know, we sort of try to see how can I construct one of these. So 40 is, okay, so this is kind of a little more tricky, so those who got the correct answer must, you know, really pretty good at doing this, or you have already figured out the general argument, so we know that if you multiply by 1, right, the product doesn't change, but, you know, we can create a whole new set of numbers, like infinitely many different numbers, because we can add just 1 everywhere. So I add 1 here, then, if I add 1 in front, it wouldn't make any difference, right, but I add 1 at the back here, I can actually make it into 5, and 2, and 2, and 2, so I really just, you know, increase the product of n a lot by adding 1 at the end, so that would give me 40, right, so 5 times 8 is 40, and that means, you know, A is also possible, and, of course, in the exam, you're going to circle 35, but at home, you're going to want to ask yourself why 35 is not possible, what is the difference when you initially, when you have five choices, is there any way we can figure out it's 35 without doing all of this? I don't know how, but, I mean, after we've gone through all of this, then we actually know why it's d, because d, if we factorize, it's going to be 5 times 7, right, and the way you get 5 is, you're going to have 4 at the end, and that is possible, we could have, you know, yeah, we probably can't even get 5 in this case, okay, we can get 5, 54, 5, but there's no way we can get 7 from adding 1 to any of these digits, right, and that means, if someone gives you 28, then you know that it's also impossible, because you have a factor of 7 in it, right, so a lot of the time, yeah, the general argument only comes up after we've actually gone through specific examples, and for number theory, I've seen that going through a specific example, so a smaller version of the same problem are very, very useful techniques. Okay, so go to the next one, the number of divisors, so if I've given a number that, you know, after, that probably have a lot of factors, how we can calculate how many factors it has, for example, given 60, right, is there any way we can, you know, find out how many factors it has without having to explicitly list out all different factors, right, so there's a theorem for that, and this is a theorem, if we factors a, this is a factorization p1 to the power e1, p2 to power e2, all the way to pn to power en, then the number of positive divisors a is a product of e1 plus 1, e2 plus 2, en plus 1, and this is actually a formula that you can actually prove it right away, because why, any divisor, it should also have form p1 times something, maybe alpha and p2, p2 times beta, then, you know, so on so forth, right, maybe I do, I'm sorry, it's right here, p1 to the n, p2 to the n, something like that, and if it doesn't have p1, then n just equal to 0, it doesn't have p2, n equals 0, it doesn't matter, I can just, you know, take all of this and put in this divisor for a, and then how many choices I would have for n, you know, it could have anything from 0, 1, all the way to e1, right, but not more than that, and there's a e1 plus 1 numbers in total, and the same thing, how many choices I would have for n would be go from 0 to e2, so that's e2 plus 1, I'm sorry, this should be e2 plus 1, and then we use the multiplication rule in combinatorics, we have a separate lesson on combinatorics, but this is where this formula comes from, and you really can actually just prove it. So, for example, we have a 60 is 2 square times 3 times 5, so in this case 1 and 1 over here, so any divisor of 60 I can write 2 to the power x, 3 to the power y, and 5 to the power z, right, and the choices of x is going to be 0, 1, 2, I have three choices, the choice for y is going to be 0 and 1, the choice for z is 0 and 1, and that's why I have this formula with 3 plus times 2 times 2 is 12, and if I list them out in pairs, then I really have 12 of them. So I hope that you understand where the formula comes from, and that would give you a lot more confidence in using them to solve questions. So let's go for it, next question. How many positive divisors are there for the number 6n if it is known that 2n has 28 positive divisors and 3n has 30 positive divisors? So this is, I would say it's a tough question. Do you have four or five minutes to do that? 2n has 28 and 3n has 30. So you can, you know, just rewrite the formula here in case you need it. Let's see how we can start with this. So, first hint, we can always start from the smallest prime numbers, right, usually I think that's all we need, they would not ask you for anything much bigger than this, 11, 13, but the idea is that because a, b, c, d could be zero, so we use algebra because sort of the most generalized form, so I'm allowed to use that, even if n does not affect two or three and five, I can still use this formula because these numbers could be zero, right, so I hope that would be the first step, and then 2n is actually 2, it definitely has a power of 2, so it's gonna be 8 plus 1, right, 3 plus b, 3 to the power of b, 5 to the power of c, 7 to the d, and so on so forth, and 3n here I would have 2 to the a times 3 to b plus 1, right, and then 5 to the c, and I just put this first step because maybe I'm sort of not new to this equation, so, but other than that I haven't really done anything substantial, except for just writing things out, okay, so I give like maybe one more minute to see if we could figure out, so the number divisor here is gonna be 28, and you write equation with 28 over here, you write equation with 30, and what would we have, okay, so it's gonna be almost everyone put out the answer, so we've used that formula, we have a, you know, this power, we have 2 plus 1, it starts from 0, b plus 1, right, times c plus 1 and d plus 1, and just, you know, so on so forth, over here I have a plus 1 times b plus 2, right, and c plus 1 and d plus 1, so at this point we'll realize that everything else be the same, I mean the cutoff is here, they're all the same, and we are only focused in the first two, right, a plus 2, b plus 1, and a plus 1, b plus 2. How to figure out a and b, because, you know, we can do the reverse thing that factorize 28 and 30, and there are not many ways to factorize them, right, because 4 and 28, of course we don't want to do 128 or 130, it would be not good, these two numbers are consecutive, so if we do that we would say that 28 is, we want numbers that are sort of close to each other, so right in the middle it's going to be 4 times 7, and 30 is 5 times 6, right, these are perfect because we have pairs of consecutive numbers over here, this one go up, this one goes down, so if we look over here, this one should be 7, right, and that should be 6, and b, that means this b plus 1 is 4, and b plus 2 is 5, right, I hope that makes sense, and everything else is going to be actually 1, so c and d are 0, that means these numbers only have factors of 2 and 3 only, and nothing with 5 and 7, so after we figure that out, then we write again n is equal to, a equal to 5, so it's going to be 2 to the 5th, and times 3b is 3, so it's going to be 3 to the 3rd, right, and that means 6n, pick up another factor of 2, 2 to the 6th, and 3 to the 4th, and that means the number of divisors is going to be 6 plus 1 times 4 plus 1, and the answer is 35. I admit that this is a difficult question, but you have a question in much, this is under a class in 99, but you have a question that in very recent years I use the same formula, and this question is really nice because it actually used, you know, brown factorization, so I hope this question is helpful. Okay, any questions, just feel free to enter the chat. You can read it at home, you can see some questions. Another question over here, a prime number is called special if it is a single digit prime number of a prime number with the property that after removing the first digit the resulting number is again a special prime, and after removing the last digit the resulting number is again a special prime, and the question asks how many special prime numbers are there. The choices are very small, so I would imagine in this case you probably would be able to create all of them, to construct all of them, right, because there's only, you know, maximum 11, so I think you can figure this question out. I'll give you a single digit prime number, first and the last, the last digit is a prime number, okay, so when I read this question I would just do, you know, three and two, three, five, seven, like that would be example, or if you have a number a, b, then a, b is a prime, a is also a prime, b is also a prime, that's what the question means. If you have a, b, c, right, then a, b, c also has to be a prime, a, b is also a prime, b, c is also a prime, that's what the question means. So this is actually a good question, you can construct all different choices, right. Question 30, so you have, I already got some answer. You probably have more time than that. I mean, you can, question 30, you probably can take three, four minutes. When you go to the three-digit numbers, remember to use the divisibility rule to check if the three-digit numbers that you constructed are actually prime. Okay, so I get started, I got some answer, but got started slowly because I think probably you figured out this part and maybe got stuck at the last one, so let's see. So the first digit is easy, right, 2357, and because a and b's are prime, so that means this two-digit number can also be constructed from 2357 only, right? Oh, great, yeah, I think we finally got the correct answer. So a, b, I know that b cannot be 2 and 5 because that would make it not a prime, so I only have a 23, 53, not 7, not 33, 7, 3 and 37, right? It's a good idea to remember all the prime numbers, two-digit prime numbers up to 100, and for sure you should know that by heart before the exam. And then over here, because it's like a, b, b, c, right, so I have to find like 2, 3, 3, 7, there's only two ways to construct, either use 3 or 7 as a digit in the middle, so I would have a 2, 3, 7, right, 2, 3, 7 is the first choice, or 5, 3, 7 is the second choice, 7, 3, 7 is the third choice, and this one, 7, 3, 7, using the 7 in the middle, so it's going to be 3, 7, 7, right? So when I was first doing it, I was like very happy because I could construct everything, but then don't forget to check, then I check like 4, 4, 4, then I would have 12 numbers, but there's no 12 here. So the trick is that we know these are prime numbers and we know that we can construct this using a, b, and b, c as the prime numbers, but we also need, don't forget to check that a, b, c itself must also be a prime number, and that's harder to check because these are two-digit numbers, so we can use our divisibility rules, right, 2, 3, 7, we know that 2, 3, 7 is, is it, yeah, 2, 3, 7 is 7, 6, okay, 2, 3 is 5 plus 7 is 12, oh my, yeah, so good, I made a mistake, so this is actually a multiple of 3, right, and then 5, 3, 7, this 8 plus 7 is 15, so we are lucky this one is also a multiple of 3, 7, okay, I'm sorry, I made a mistake here, 7, 3, so it should be 3, 7, 3, 7 is not divisible by 2, not divisible by 3, not by 5, definitely by 11, so we know that 7 plus 7 is 14, 14 minus 3, so we are lucky again, this one is exactly 11, so this number is divisible by 11, so it's also not a prime, 3, 7, 3, the square root of that is close to 20, so we have to check all the prime numbers up to 19, so we do 7, 11 is not, 13 is not, with 13, 9, because we have 53 is not, so 17, 19, actually this one is the only prime numbers among these choices, right, so we have a 3, 7, 3, and that's the three digits, but we have only one three-digit numbers, and from this we cannot construct four-digit numbers, right, and I think that would mean that we cannot stop here, so we have a 4, 4, and 1, that's number d, right, so again this equation, you can definitely do, the only thing that's sort of the limiting factor here is time, right, because it takes a lot of time to not make a mistake in constructing this and be able to figure out quickly which one of these is the prime numbers. Okay, so any questions about this? Okay, so yeah, I sort of just want to recap everything, you sort of, we went to maybe six, seven questions, but as you can see that a lot of questions when you first read it and you have no idea of how to process, but really for I think maybe more than half of the questions in this class we could do it by going to a specific example, by eliminating obviously like sort of easier choices that are wrong and narrowed out our choices, and really start by, you know, doing slowly prime factorism using algebra is definitely, as you can see that we work with very small numbers over here, but algebra is really helpful, a lot of questions we don't even know how to start without writing the numbers in generalized forms, and as I said, emphasize earlier, really one of the biggest help in getting and in solving number theory problems in a short amount of time that you will have to good at arithmetic, so make sure you know all the numbers up to a few hundreds prime factorizations. Any other questions? So next week we're going to do sequences and patterns, and these are actually some of the very most interesting question in math and group falls in these categories, and by sequences and pattern we do not mean just like arithmetic sequence or geometric sequence, but, you know, a lot of other sequences on transformations, and I think the questions and I think the questions are sort of more interesting, but maybe less difficult than number theory, so I will stop here for the lesson, but we will stay, so anyone has any questions for today's class or previous one, feel free to ask us.

number theory

Diophantine equations

prime numbers

positive divisors

prime factorization

divisibility rules

factorization techniques

algebraic identities

difference of squares

sequences and patterns