ok so here you have a topic called factors and multiples so what is I mean very special about this topic. All of us know about factors and multiples. But the interesting part is about the questioning and answering because we are going to come across some interesting questions in factors and multiples. But anyhow, I am going to tell you the introduction part because we need to know the concept clearly.
When a natural number n divides another natural number m exactly into natural number of times. So, n is a natural number. And m is also a natural number.
Another natural number could be itself also times. Then m is said to be factor of m. m is said to be multiple of m.
So factor multiple comes at the same time. So here you see 4 divides 36 exactly in the natural number of times which is 9. So if you divide 36 by 4, it goes 9 times. Get a reminder. So, therefore 4 is a factor of 36, so 4 is a factor of 36 and simultaneously we can say 36 is a multiple of 4 and this concept of factors and multiples takes for only natural numbers.
So, the concept of factors and multiples. So, it is only for, holds for natural numbers. One question is, if suppose I take 5 and divide by 2, say there are 5 apples you need to divide to 2 persons, you can say 2 and a half or 2.5 that goes exactly 0. That means that 2 divides 5 exactly, no.
So, 2 is natural. 5 is natural, but 2 and half is not natural. So, we need to check the natural number should divide another natural number exactly into natural number of times.
That is the first sentence you can see there. So, that means all should be natural. The remainder should be 0. Then only we can talk about the divisor is a factor of the dividend and the quotient should also be a natural number. So that is the idea and conversely we can say dividend is a multiple of divisor.
The quotient is also a... natural number because you should not look at the remainder exactly means remainder should be 0, but exactly should be a natural number of times. So, this is not going to work the concept of factor on multiples for the simple reason 2 divides 5 exactly but not into natural number of times, it divides into a non-natural number of times. So, that is the thing about the factor on multiple that we already know that just want to emphasize on that.
concept. So, factor multiple will not be applicable for fractions or anything. It will be applicable only for natural numbers.
So, let us look at the properties. Now, because we are going to have some interesting problems on factors and multiples. First, a natural number is limited finite number of factors.
So, any natural number because any number above, I mean if you take a number like 10, above 10 cannot divide 10 exactly into natural number of times. We can say 20 into half is 10, but half is not a natural number. So, whatever number you take natural number, you can start from 1 till that number.
So, it will have only limited number of factors. For example, if you take 10, you can check only from 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 for the factors. So, this is the number.
So, you want to check for factor, factor of 10. So, you are going to have, from the limited lot only you are going to check. So, obviously it will have limited number of factors. You are going to get yes or no.
So, 1 is a factor of 10 because it divides 10 exactly 10 times. 2 is a factor of 10 because it divides 10 exactly 5 times. 3 is not a factor. It divides 10 exactly by 3, 1 by 3 times.
It is a fraction. 4 does not divide them exactly into natural number of times, 5 divides 6, no, 7, no, 8, no, 10, yes. So, there are 4 factors.
So whatever natural number, even if you take big like 2022, from 1 to 2022, you theoretically check, you might get tick and intos. Even you will not get all ticks, even if you get all ticks, it is finite only because you are going to check from 1 till that number. So therefore, a natural number has finite number of factors. A natural number has unlimited infinitely many number of multiples. 10 divides how many numbers or what are the multiples.
So, if you take this number 10, what are the multiples? Multiples means you again concentrate from what natural numbers. In fact, 0, 10 zeros or 0 can be taken as a whole number multiple, but multiple concept will come only from natural.
So, 10, 20, 30, 40, 50 and where it ends, never ending. So infinitely many number of multiples, unlimited. So that's the difference.
If you take any natural number, it has finite number of factors or it has infinite number of multiplications. If you take prime number, it will have only two factors. A prime number has only two factors.
What about 1? 1 has only one factor. And what about composite? It has more than two factors.
So, any natural number that has exactly two factors is called as prime number. Any natural number that has more than two factors is called as a composite number. And 1 is neither prime nor composite because it has only one factor.
Two or more natural numbers have at least one common factor because one is definitely a common factor. Every number highest among them is called highest common factor HCF. Two or more natural numbers have infinitely many common multiples. The least among them is called LCM. So if you take for example 6 and 8, what are the factors of 6?
1, 2, 3 and 6. What are the factors of 8? 1, 2, 4 and 8. So, what are the common factors? Common means it is a factor of both. 6 and 8 are 1 and 2, that is all.
3 is not a common, 3 is a factor of 6 but not 8. 4 is a factor of 8 but not 6. So, it is not a common factor. 6 is a factor of 6 but not 8. 8 is a factor of 8 but not 6. So, 1 is a factor of 6 as well as 8. So, it is a common factor. 2 is a factor of 6 as well as 8. So, the highest common factor, let us see f of 6 and 8. So, among the two which is the highest 2 is 2. So, highest can be only one. So, this is about highest common factor. So, if you go for the multiples, multiples of 6. multiples of 8, never ends.
So if you look at the common multiples of 6 and 8, what are the common multiples? 24. So, that is there, that is the first where it meets. Then next to 24 it meets, just like the clock, minute hand and hour hand meets at 12, 12, 12, again it meets at next 12, 12 like that.
So, it meets another next to 24. So, the 30 acts like 24 plus 6, 36 is 24 plus 2. So, it is like the next 6, next to 12 like that. So, this is the next year 8, 24 plus 8. 24 plus 16, 24 plus 24. So, as you can see that 24 plus 24 will come in the multiple of 6 also. So, every 24 they meet.
So, that means there will be infinitely many multiples, but among them the least common multiple, who is the least, that is 24. There are methods to find it, but you should understand what is the meaning of the least common multiple or highest common factor. So, there is something called common multiple, then only. First multiple, then common multiple, then least common multiple.
Same thing factor, common factor, then highest common factor. But the common concept can come only if you take two or more. You cannot talk common only with one fellow.
You can only if you can talk taller among the two or shorter means. You need at least two or more to compare. So, this is what is about factors and multiples and I have not written but I have told you earlier that please you can make a note.
Prime numbers are exactly two factors, composite numbers are more than two factors and one is neither prime nor composite because it has only one factor. So, it will not fall in the category of primes, it will not fall in the category of composite numbers. So, we all know all these things but I just want to emphasize the basics. But when it comes to application on problems.
We must have come across finding HCF and LCM by some methods set in normal school curriculum. But we are not going to the problems that way. Our problems will be something different.
It will be even testing the basics and also some techniques, applications. Some interesting problems will come across. So that we have problem set 1, 2 and more problems. So let us look at problem set 1. What are the two digit common multiples of 6 and 8? Two digit.
So, we know that common multiples of 6 and 8 just now we saw. The common multiples of 6 and 8 are 24, 48, 2, 96, 120, 144, etc. But you want only the two digit.
Therefore, the answer is. What is the answer? The two digit common multiples are, sometimes you might come across this type or 24, 48, 72 and 96. So, overall least common multiple is 24. Therefore, we have been doing after LCMHCF and all assuming that we know about multiple common multiples. Then only you can go to least common multiple. But when I come across some students, they really about common multiples or common factors, they are not able to answer because they directly do some methodological way of finding LCM.
So, they should know what is actually LCM. So, for CM only they should know there are lot of CMs among which there is a least called LCM. There are lot of CFs, common factors among which we get highest common factors.
So, that way we should learn the idea. So, if the question is after how many? 4 digit common multiples of 6 and 8 are there.
It is indirectly asking how many 4 digit multiples of 24 are there because these are all multiples of 24. The common multiples of 6 and 8 are multiples of 24. So the question is indirectly, this question is indirectly equivalent to how many 2 digit multiples of 24 are there. Multiples of 24 are there. Questions like that.
This is equivalent. What about the next question? The next question is what are the two digits?
So, what is the highest common factor of 600 and 800? It is 200. 200 is the common factor, highest common factor. How do you know?
as well as 800. There won't be any number that will divide, so there is no bigger number than 200. So, it is the highest common factor. So, hcf of 600 and 800 is 200. If there is any number above 200 divide 600, then for example, suppose 220 divide 600, then next multiple of 600 will be 600 plus 220, that is 820, it can't divide 800. So, if 200 divided by 600, 600 plus 200 is also 800. So, 200 is the highest. Any number above 200 cannot be a factor of both 600 and 800 because then 600 should be a multiple of that number as 800 also.
But the gap is only 200. So, you cannot have a factor more than 200. So, that is the idea. So, then this question, what are the two-digit common factors? So, any factor of 600. sorry, any factor of 200, sorry, is also a factor of 600 and 800 because 200 divides 600. So, anything divides 200, you divide 600 directly.
So, it is a simple theory. Therefore, what are the two, so the question becomes not how many actually, what are they? So, I wrote how many, actually the what are they. The first question is what are they?
The second question is also What are the two digit? What are they is different from whom? What are the two digit factors of 200?
What are the two digit factors of 200? So the two digit factors of 200 are, so for the second part the answer is, so you have 50. 50. Right. 50 is there, 40 is there, 40 into 5, 50 into 4 and then 25 into 8, 20 into 10. So these are the two digit factors. So that is all. I think I made a mistake by writing how many, it should be what are they.
So what are they means you have to list, how many means you need to count and tell them. So when it comes across in a test. objective test, you will never ask what are they, you will be asking only how many. Let us move on to the, so this is a basic question, can the HCF and LCM of a set of natural numbers be 4 and 125 respectively, give reasons.
Can the LCM and, HCF and LCM of a set of natural numbers be 4 and 126 respectively. something like A1, A2, A3, let there be AN, set of N numbers, let the HCF be 4, let the LCM be 125. So, what is the connection? 125 is a multiple of a1. First of all, LCM means it should be a multiple of a1, a2, a3, common multiple. It is the least common multiple of all.
That means 125 is multiple of every number in the set. That is, let us say a1. Then 4 is is2 common factor of a1, a2 and all. That means conversely a1 is a multiple of 4. If 4 is a factor of a1.
a1 is a multiple of 4. That means 125 is a multiple of a1, a1 is a multiple of 4. These two put together, you get 125 is a multiple of 4, which is false. Therefore, the answer is no. So, indirectly we have a property, LCM of any set of numbers. should be a multiple of HCF. If not a multiple of HCF, there is no set.
So, LCM of a set of natural numbers should be a multiple of its HCF. So, this is a very important property. So, if you look at 1, 26 and 4, 1, 26 is again is not a multiple of 4. So, with the similar argument, so we get no, that is also false. So, this is also no, there is no set. Because if 4 is a factor of one number, the set and that number in turn is a factor of 1, 26, just do the converse.
If we say i is to common factor means it should be a common factor, it will be a factor of any number. Conversely, that is a multiple. 4 is a factor of a1 means a1 is a factor of 4. 4 is a factor of a2 means a2 is a factor of 4. So, you take the converse now.
So, directly you get 126 is a multiple of a1 which in turn is a multiple of 4. That is not true because 126 is not a multiple of 4. So, LCM of any set of natural numbers should be a multiple of its LCM. This is a property. Let us move on to the next one.
Find all two digit numbers. 9 less than a less than b less than c less than 61 where hcf of bc equals to a and lcm of ab equal to c. This is a different kind of problem.
We are all testing the basics. We are into the problem set 1, remember. Now, highest common factor of bc is equal to a. We would have come across in English like a What is that? That is you are trying to write the sentence in two different ways, subject to object you just invert like Rama killed Ravana, Ravana was killed by Rama something like that.
So, if n is a factor of m. M is a multiple of A. So, that converse only the trick you have to do. First of all, BC, HCF of BC is A means B and C are multiples of A.
Then only it will be highest complex. So, therefore, B is equal to something like AP and C is equal to AQ, but P is less than Therefore, ap is less than aq that means p is less than q. So now this sentence we have assumed said b equals to this. LCM of ab is equal to c. What does that mean?
So this is another interesting part. This tells you again c is a multiple of a which we already know because this is common with the group. as well as B.
So that means A will be A, B will be something like AP and C will be something like APR then Q must be equal to PR. because it should be a multiple of B. So, this is how you will get.
Now, coming back to that HCF of B and C. So, what will be HCF of B and C? AP is there and APR is there.
So, at least AP should be the HCF or what is the LCM of A and B? What is LCM of A and AP? It will be AP.
That means that is B, not C because C is different from B. So, therefore, that HCF of BC is equal to LCM of AB equals to C is not working that becomes B, not C, but it is given as C. Similarly, if you look at HCF of B and C, you will be getting B, not A because HCF of AP comma APR, that will be equal to AP. is equal to b not a, unless p equals 1. If p equals 1, a and b will be equal, but a is less than b. And this is just to test you the basics.
Therefore, there is no such ripple a, b, c, not only 9 less than a less than b less than c less than 6, not only two digit numbers. When generally you take any Natural numbers a less than b less than c. Forget about that 9 less than 61. That is simply to divert your attention because we will start searching them.
Sometimes we will go for searching from 10 to 16 and wasting our energy. For that only it is trying to divert that question has been done. So, if you go by the theory, you can see that there exists no such triple because it contradicts.
That HCF of BC is a and LCM of ABC will never happen if all are distinct. all are distinct. So, therefore, this is what you should understand.
Fine, let us move on to the next problem. The product of two natural numbers is always equal to the product of the HCF and LCM. This is for two natural numbers.
This is a very famous thing. But I am going to show you the proof. See, that means we are concentrating on the theory, the problem set one.
Let me go back and tell you. First one is just the basics of understanding that multiple, common multiple and then what, two digit common multiples, not least common multiple. Same way it is about the common factors.
And then the second one is we got a property that LCM of set of natural numbers should be a multiple of HCF. The third one is if you have three distinct numbers ABC, and the HCF of the larger two is smallest. or the LCM of the smaller to its largest can never happen, you cannot find such a typical.
These are all some properties that if they are all distinct. Then the fourth one we are going to see this is a very famous result. But we generally we start applying this in the school curriculum, but I do not see a proof.
We need to prove it theoretically. Let A, B be the two natural numbers. whose HCF is H.
So, therefore, we can keep A to be HM and B to be HN where M, N are co primes. What do you mean by co primes? They don't have anything common other than one. So, otherwise what will happen if m and n has something common 2 suppose you take, then a will be h into 2 into something and b will be h into 2 into something, I mean whatever be the other part. But then h into 2 will become the highest common factor, not h.
So, m and n should not have common factor other than 1. Co-primes means they have common factor only 1, their common factor is 1. common factors and that is also the highest common factor because it is only one. I am talking about the m and m. What about the LCM? Therefore, LCM of AB will be HMn, least common multiple.
HM into n is HMn, Hn into m is Hnm which is Hmn. So that is a multiple of both A and B. That is the least common multiple.
Don't write HM into Hn. H squared mn is also a multiple but it is not the least. So now what is hcf into lcm?
hcf into lcm is equal to h into hmn. That you can interchange and rewrite as hm into hn. That is equal to a into b. So hence it is good.
But for three numbers this is not going to work like this. So, that is why we need a proof, need to write definitions in the form. So, this is a simple theoretical way of writing.
Of course, you have to write that Mn or co-primes like that, LCM is HM. Then you verify HCF into LCM is H into HMn, that is HM into Hn. a into b, n-set is proof. Let us move on to the next one. Can the HCF of this such a big number, 1, 2, 3, 1, 1, 3, 1, 3, 5, 1, 9, 3, 8, 3, 4, 1, 1, 2, 3, 1, 1, 3, 1, 3, 5, 1, 9, 3, 8, 1, 7, 4, 1, 4, be 18. So, they will be doing like the school method, this box and all, when will you finish?
It will be difficult. Therefore, the idea is be 18. Suppose they are multiples of 18. Let us assume that hcf is 18. See, this is called proving by contradiction. Suppose it is not. Let 18 be the highest common factor or even you do not require the highest common factor, be a factor, common factor.
Let 18 be a common factor of. Let me call this A and B. Let me call this as A, this fellow as B. So, A is a multiple of 18. So, reverse, B is a multiple of 18. I can write b in the top or a in the bottom. b is bigger than a, so I am going to do something.
So if I assume then that implies b is a multiple of 18, a is a multiple of 18. This implies b minus a is also a multiple of 18. So that is the catch. So if 18 is a common fact, first of all, Forget about IS-convector, it should be a common factor. Among the common factor, it should be IST.
If it is not a common factor, then it cannot be IS-convector also. That's the point. So, 18 is a common factor of A and B.
It's a factor of both A and B. That means B is a multiple of 18, A is a multiple of 18. B minus A. What happens? The starting digits 1, 2, 3, 1, 1, 3, 1, 3, 5, 1, 9, 8, 1, 3 till 1. They get cancelled.
you can see till one it is the same, so you get cancer, this block gets cancer. Then it is like you are doing only 7, 4, 1, 4 minus 3, 4, 1. All others the block, you know, they get cancelled because it is the same. You get something like 4000. 4000 is a multiple of 18. It is not true because 18 is multiple of 9 and 2. 4000 is not. Some of the digits.
So, this is not true. Contradiction, what is mean by contradiction? The starting fellow only gives us something false. All the other process in between correct. If you assume 18 to be a factor of a and b, b will be multiple of 18, a will be multiple of 18, b minus a is a multiple of 18, 4000 is a multiple of 18, but 4000 is a multiple of 18 is not true.
Therefore, that starting point is suppose if you get some multiple of 18, it may be true only will come. Please try to understand that also. But if it is not going to happen, this is not true. The starting input only is not true.
So it's like a, maths is like a device. You feel some, like a machine when you put an input, you get an output. So the output if it is a faulty one, I am talking about a mechanical device. If you put an input, you get a faulty output. And fault might be because of the input.
or because of the process that machine, machine might have gone wrong due to some wear and tear. So, only two options are there, either the input raw metal is not correct or the processor is not correct. So, you get a wrong output.
But in mathematics, it is a theory machine, you are putting the theory, everything is correct. Machine is like 100% efficient, no wear and tear. That means input is only the error. What is the input? 18 is a common factor.
So, 18 is not a common factor. That is what the contradiction tells you. So, 18 is not a common factor of A and B. What does that mean?
Therefore, 18 cannot be HCF. So, no number which is not a factor of 4000 can be a common factor. That is very important. The highest common factor cannot be more than 4000 also. First of all, the 10 and all will not come.
So we can actually get rid of lot of things. I will be showing some different methods of finding highest common factor like that because it is not like the traditional one we can do. So this particular problem such two big numbers and we are asking whether 18 is the highest common factor. So, let's go. I mean you can just go with the basics.
Let us take 18 to be a common factor then just like I told you the active and passive like that you just revert it. If 18 is a factor of A and B, B is a multiple of 18 and A is a multiple of 18 then what happens B minus A is a multiple of 18 then starting initial digits will get cancelled if 4000 is a multiple of 18 which is not working therefore 18 cannot be a common factor so 18 cannot be a highest common factor as well. Let us move on to the next one. If HCF of two natural numbers is equal to one-third and one-fourth of the two natural numbers, HCF of two natural numbers is equal to one-third and one-fourth of two natural numbers, HCF is less than 9 and LCM is greater than 85, you find the two natural numbers. So, this is not a tough problem.
The problem how do we approach, no trial and error actually. Let me tell you, this one-third and one-fourth Hcf is a natural number, it cannot be a fraction. Therefore, the two numbers must be multiple of 3 and multiple of 4. Let the two numbers be 3H and 4H.
Let the Hcf be two natural numbers. The main natural numbers only. We can talk about the concept of factors and all per natural numbers, 3H. So the two natural numbers must be 3H and 4H. So do not write X and Y and write X by 3, Y by 4, X by 3 equal to Y by 4, all those things you do not require.
It is better we define 3H and 4H. So 3H does not mean it is a multiple of 3. If H is something like 2 by 3, 3H will be 2. But here highest common factor, but generally I am telling, you cannot assume 3x means multiple of 3 like that. So any unknown we have the power to define as 3h or h or 4h like that. But in this case, h is a natural number.
So 1 third of another means that should be a multiple of 3 times. It is converse just like it would active and passive only the game here. If one number is 1 third of another, HCF is 1 third of one number, that number the, that is 3 times HCF. If HCF is one fourth of the second number, the second number will be 4 times HCF that is all. So, define H and then do this.
Now, what is your H is less than 9 that is given. H is less than 9 given. Also given LCM of 3H comma 4H.
What is LCM of 3H and 4H? 12H, that is the LCM. So LCM of 3H, 4H is 12H.
So 3 into H, 4 into H, 3 into 4 into H. So H will be greater than 85 by 12. H is greater than 84 by 12 plus 1 by 12. So it is greater than 7, 1 by 12. So, h should be greater than or equal to 8, h cannot be even 7, it is more than 7, 1 by 12, but h is an integer, natural number, so it should be more than 8 or more, but h is less than 9. So, h is less than 9, but 8 or more, but it is a natural number. So, what do you say about h? Therefore, h must be 8. If h is 8, what do you say about the numbers? The numbers are 3h equals to 3 times 8, 24 and 4h equals to 4 times 8, 32. So, the numbers are 24 and 32. So, these are all very simple problems but slightly variant.
It is not like the usual problems we get. So, fine. Let us move on to the next problem. ABC are all two digit numbers, each less than 50 such that HCF of AB equals to 2, HCF of BC equals to 3, HCF of CA equals to 5. Find all combinations of such triples ABC.
This looks like an easy problem. It is not. You have to really find it.
Of course, if I thought of giving all two digit numbers. Sorry, not each less than 50. Then we will be getting something more. So, count will be bigger. Let us see.
What will be the form of A? What will be the form of A? Form of A will be 2 into, because 2 is there. B also 2 into.
Because HCF of A and B is 2. There is a 2 definite common factor. But 2 into whatever you put here. and whatever you put here, those two should not have anything common other than 1. Otherwise, that hcf of a and b will not be. So, there is a 2 into.
So, what about c? b and c has a 3, that means there is a 3 here, there is a 3 here, 3 into, that means a should not have 3 in its, I mean factorization, 3 is not a factor of a. And then what about hcf of c and a is 5, that means there must be a 5 in a, 5 minus.
So, if I am going to, I can call this fellow, this could be a, we are not doing a prime factorization, remember this, the balance box, whatever it is, it could be a composite that can be broken into primes. So, let us take these three fellows, this is something like asterisk, dollar and tick, that is the balance. What are the characters of asterisk, dollar, and t?
Suppose I fill start A and then B and then C. Suppose we fill A, then B, then C. Let us take like this.
So, that means we have to look on asterisk. What do you say about the asterisk? Asterisk should not be 3. It cannot be 3. Why? and remember all are two digit numbers this is a b c or all So, 2 into 5 into asterisk cannot be 3, why do you say it is a, because b has a 3 here, 2 into 3, so if you put a 3, a and b already have a common factor 2, highest common factor, so then highest common factor will be 2 into 3, that will not, that will contradict with ZCF of a and b to be 2, so we should maintain this, so asterisk cannot be 3. be 3 and what is the maximum for asterisk cannot be 10 or more, why?
Cannot be 3, it cannot be 10 or more, asterisk cannot be 10 or more, why it cannot be 10 or more? Because 10 into 10 is a 3 digit number, so some combinations for asterisk, this is sorry. We call this something like 1, then 2, let us concentrate on dollar.
What do you say about dollar cannot be 5? Why? If dollar is 5, then the common factor of A and B will be 2. There will be 5 in A as well as 5 in B. There is a 2 already.
So, again same thing with B and C. B and C's common factor is 3. So, if you have a 5, then there will be 3 into 5. That will become common for. So, dollar cannot be 5 and dollar cannot be asterisk or its multiple and dollar cannot be 17 or more.
That is the third tick. Tick cannot be 2, like C and A is common, should be 5. Tick cannot be asterisk or its multiple. Tick cannot be dollar or its multiple or even you can say the factor. us factors we can use because if asterisk is bigger then we might be using.
So, wherever I use the multiple you can even use the replace by factor means one will be a factor then one has got a special case we are talking about that is why I put multiple a factor that is not one we can take replace like that. Tick cannot be 7 or more. because 15 into 7. So, these are the facts that you should think and you need to work on that.
So, if you are going to try that, let me quickly put a table and then try to get the result. Asterisk, dollar and tick. Correspondingly, we can have A, B and C.
Asterisk we can have 1, dollar we can have 2. 2 and tick we can have 1. So, 2 into 5 into 1 that will be 10, 2 into 3 into 2 that will be 12 and this will be 15. So, this is one solution. So, these are the solution triples. So, what we have is the solution triples, find all possible, this is 1. Then the next one is 1, 2 and 3 that is 10. 12 and 45. Okay, you can verify. Okay, so can you try? So just pause on and then the pause the video and try to get all the combinations but remember the conditions.
But I have written just for you to understand but while working should be careful right. If I put asterisks, that means asterisks is 1, a will be 10, 2 into 5. So, you need to carefully put b. Of course, dollar cannot be 5 and tick cannot be dollar or it cannot be 5. And then you have to carefully do. Then once asterisk 1 is gone, then we should go for asterisk either 2 or I mean 2. Asterisk cannot be 3, remember.
Then we can go for asterisk to be 4. Try to understand that. Asterisk cannot be even 6 because It cannot be 3 or multiple of 3 also. You understand?
So, cannot be 3 means it can't be multiple of 3. So, you will have only less values and the asterisk is going to be 10 also, so asterisk is going to be 9 also because, so you will be getting only few values to be tested and then you will be getting the combinations which you may have to try quickly. So, you pause the video and try, I mean the combinations and then I am going to show you the combinations, you can verify that but then you pause and then give it a try. So these are the combinations of asterisk, dollar and tick that is 2 into 5 into asterisk is your A, 2 into 3 into dollar is your B, 3 into 5 into tick is your C.
And then you can see that. because we want to find all possible combinations means we should not know how many combinations I made the mistake in the beginning like the difference between what and how many and then I corrected it so you need to find means you need to list out ABC for that you need to check the asterisk, dollar and tick and then each value should be less than 50, remember that also Each value is less than 50. So, we need to get only the combinations of ABC. Each of them is less than 50, but all are two digit numbers.
That is also the condition. HCF of AB is 2, HCF of BC is 3, HCF of C is 5. So, these are the combinations 10, 12, 15, 10, 12, 45, 10, 18, 15, 10, 24, 15, 10, 24, 45. 10, 36, 15, 10, 42, 15, 10, 42, 45, 10, 48, 15, 10, 48, 45, 20, 18, 15, 20, 42, 15, 20, 42, 45, 40, 18, 15, 40, 42, 15, 40, 42, 45. So, if you take any, for example, you take 10, 42 and 45, 10 and 42 is the highest common factor is 2, 2 divides 10, 2 divides 42, there is nothing else. 5 divides 10 but 5 doesn't divide 42. 10 divides 10, if you take 3 divides 42 but then doesn't divide 2. So, if you take 10 and 45, the highest common factor is 5. If you take 42 and 45, 3 is the highest common factor. So, therefore, if you take any of the 16 solutions, the A, B, C, A and B's highest common factor is 2 and B and C's highest common factor is C, C and A's highest common factor is 5. And these are all. the combinations of triple CBC, you have 16 solutions.
Just imagine if you slightly increase it to 100, then there will be plenty, it will be not double, I do not think it will double, it will go 4 times in 60 something, 60 solutions. So, I mean if this is given something like 5, how many like that, you have to be little careful. Let us move on to the last problem in this problem set 1. H, A, B and L are all two digit numbers such that HCF of AB equal to H that is less than A, that is less than B, that is less than L equal to LCM of A.
That means all of these are two digit numbers. Find all possible combinations of such pairs A and B. Remember that AB is HCF is H, AB is HCF that is H, LCM is this. So all the four are different in values.
Which is lesser than? smaller number, that in turn is lesser than the larger number, that in turn is lesser than the LCM. Because sometimes if you take two numbers for example 10 and 20, HCF is 10 and LCM is 20, if one number is multiple of another, two numbers you take, the bigger number is multiple of the smaller number, smaller will become the HCF and bigger will be the LCM.
But that is not going to happen here. So therefore, B is not a multiple of A. it is a multiple of A it will happen, so B is not a multiple of A, it is not a, but we want all two digit, so how do we go, A should be multiple of H, B should be multiple of H, so let A equals to HM, B equals to HN.
So, a less than b implies hm less than hm, that means m is less than, what is the LCM? So, h is equal to h. it is HCF, and L is equal to HMN, remember N is not multiple of M, so B is not a multiple of A, N is not a multiple of M. So, now L is equal to HMN.
So, H, These are all two digit numbers. In fact, h can go till 10, 10 is less than, so we have 10 less than or equal to h. So, it is very interesting now. Now, I have a very interesting spec that 10 less than or equal to H that means 10mn is less than or equal to Hmn.
I can write like that. 10 less than or equal to h, multiply m and both sides, but hmn is a 2 digit number less than 100. Therefore, 10mn is less than 100. That means mn is less than 10. So m into n should be maximum 9. But m is not 1. If m is 1, a will be equal to h. m and n are not 1. So, that is also important, n is not a multiple of m, m should be more than, both of them are more than 1, n is greater than m is greater than 1. So, what are the combinations of mn that will be less than or equal to 9?
Let us write the combination of nm, the combination of nm, sorry mn, m is smaller, let me write. What are the combinations that works? Both of them cannot be also. 2, 2 is not working. Remember, n is not a multiple of m.
So, 2, 3 works. 2, 5. No, m, n should be less than or equal to 9. It should be less than 10. 3, 3. No, not possible. This is the only combination that will work.
Interesting. If m is 3, n should be more than m. So, 4. 3 into 4 will be 12. So, m cannot take 3, m can take only 2. But if m takes 2, n cannot be 4, n should be more than m. Because 2 and 4, if n takes 4, n takes 4 and n takes 2, 2h and 4h.
So, b will be a multiple of a, then LCM of a and b will be b. So, that is where we go by the characters. And then finally, we got m should be 2 and n should be 3. That means, me, of course we got something interesting. Finally, we have m to be 2 and m to be, let me rub all these, erase these things and rewrite.
So, we got m to be 2 and m to be 3. It is going to be 3. So, what does that mean? Our numbers will be H is then 2H is then 3H and the LCM will be 6H, all two digit numbers because HM, HN, we got HM to be H into 2, 2H, HN will be 3H, that combination only works. So, these are all two digit numbers. So, let us try to look at the combination 10, 20, 30, 60, it works 20 and 30 is LCM is 60. So, if I take 11 then 22, 33, 66. So, if I take 12, 24, 34, sorry 36. 6 into 12 is 72. So, it goes till 16. 17 it does not work because 17 into 6 is gone. Therefore, what are the pairs of AB?
So, the pairs are, pairs AB are, we want AB, the two numbers. 20, 30, 22, 33, 24, 36, 26, 39, 28, 42, 30, 45, 32, 48, 34, 51. I think I made a mistake because 17 into 6 will not work, right. It stops with 30 to 48 and 30 to 48, correct because 34, 51 if you take, so for example, 34, 51 will happen.
1D if you get 17 as H, but that 17 into 602 is not a 2-digit pair. So we get 1D, these are the pairs, 7 pairs. The question is how many pairs?
There will be 7. So you might be getting third type of question in objective. For which first, if you go by trial, the problem is like you may not be going. So you need to know the theory also, the basics, fundamentals. So, in this particular problem set, you would have seen that the theory is also important and finally we have come across some problems. I mean the problem solving is based on the theory.
We are calling it as HM, HN, HMN and then you are looking something like I put that Mn to be less than 10 and then I found the only combination works is 2 and 3 and then I went. Sometimes you go by trial and error, you might still get that 2 and 3 but you might still have a doubt. But if there is another combination of M and N, suppose the problem is like that, you might be missing that. That is why it is better to go by the theory and be sure that A and B are 2 times the highest common factor and 3 times the highest common factor only for this particular configuration of two-digit numbers.
All are two-digit numbers and all are distinct. So that can work only if the two numbers are. 2 times highest common factor and 3 times highest common factor and the LCM is 6 times highest common factor. So, 16 is highest common factor is less than 100 means. Highest common factor should be less than 100 by 6. So, it should be less than 16, 2, 2 third, 16 plus 2 by 3. So, therefore, but it is a natural number.
So, it can go from 10 to 16. Sometimes we need not write everything. Here since it is asked to find all combinations, I have written. But otherwise, how many means we may have to go after the count.
Let us look at into the problem set 2 and more problems, very interesting problems we have in fact some multiples. We will see those problems in the problem set 2 and more problems. That is all for the introduction of problem set 1.