Finding Common Ground

Hello my dear students! Welcome to today's mathematics lesson. I am so happy to see you all here, ready to learn something new and exciting. Today, we are going to study Chapter 3, and we call it "Finding Common Ground". This is such an interesting chapter because it deals with two very important concepts in mathematics - the Highest Common Factor, which we call HCF, and the Lowest Common Multiple, which we call LCM. These are ideas that you will use throughout your life, not just in mathematics, but in many real-life situations too. So pay attention, and we will build this chapter step by step, from simple ideas to more complex ones. Are you ready? Let us begin!

Let us start with a story. Imagine a girl named Sameeksha who is building her new house. The main room of her house is 12 feet by 16 feet. She wants to cover the floor with square tiles of the same size, and she wants to use as few tiles as possible. Also, she wants the length of each tile to be a whole number of feet - no fractions, just whole numbers like 1, 2, 3, and so on. What size tiles should she buy? This is a very practical problem, isn't it?

Let us think about this carefully. The breadth of the room is 12 feet. For the tiles to fit perfectly along the breadth, the side of each tile must be a factor of 12. Do you remember what factors are? Factors of a number are the numbers that divide it completely without leaving any remainder. Similarly, for the tiles to fit along the length of 16 feet, the side of the tile must be a factor of 16. So, the side of the tile should be a factor of both 12 and 16 - it should be a common factor of both numbers.

Now, let us list the factors of 12. The factors of 12 are 1, 2, 3, 4, 6, and 12. These are all the numbers that divide 12 completely. Now, let us list the factors of 16. The factors of 16 are 1, 2, 4, 8, and 16. Now, what are the common factors? These are the factors that appear in both lists. The common factors of 12 and 16 are 1, 2, and 4.

So, Sameeksha can use square tiles with sides of 1 foot, 2 feet, or 4 feet. But which one should she choose? She wants to use as few tiles as possible. If she uses 1-foot tiles, she will need many, many tiles to cover the entire floor. If she uses 2-foot tiles, she will need fewer tiles. And if she uses 4-foot tiles, she will need the fewest tiles of all. So, she should use the largest sized square tile, which is 4 feet. That makes sense, doesn't it? Using larger tiles means fewer tiles are needed, and that saves money and effort.

So, the largest square tile she can use is 4 feet by 4 feet. This is a perfect example of finding the Highest Common Factor. The Highest Common Factor, or HCF, of two or more numbers is the highest, or greatest, of their common factors. It is also known as the Greatest Common Divisor, or GCD. In this case, 4 is the HCF of 12 and 16.

Now, let us think about another situation. Suppose Sameeksha did not insist that the tile length must be a whole number of feet. What if she could use tiles of any size, even fractional sizes like half a foot or one and a half feet? Would the answer change? Well, if we allow fractional lengths, then theoretically we could use even larger tiles, like a tile that is exactly the size of the room itself - 12 feet by 16 feet is not square though. But if the room were square, say 12 feet by 12 feet, we could use a single 12-foot tile. But that would not be a square tile that fits perfectly in both dimensions if we allow fractions. Actually, if we allow any fractional length, we could use a tile of size 12 feet by 12 feet, but that would only work if the room is square. For a rectangular room like 12 by 16, we cannot use a square tile that covers both dimensions exactly unless we cut it. So, the whole number requirement is important here. This is why we need to find common factors - only whole numbers work for this problem.

Now, let us look at another example. There is a merchant named Lekhana who purchases rice from two farms. She bought 84 kilograms of rice from one farm and 108 kilograms from the other farm. She wants to pack the rice into bags, with each bag containing rice from only one farm, and all bags must have the same weight. The weight of each bag should be a whole number of kilograms. She wants to use as few bags as possible. What should be the weight of each bag?

This is similar to the previous problem. To divide 84 kg of rice into bags of equal weight, the weight of each bag must be a factor of 84. Similarly, to divide 108 kg into bags of equal weight, the weight of each bag must be a factor of 108. And since she wants to use bags of the same weight for both farms, the weight must be a common factor of both 84 and 108.

Let us list the factors of 84. The factors of 84 are 1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42, and 84. Now, let us list the factors of 108. The factors of 108 are 1, 2, 3, 4, 6, 9, 12, 18, 27, 36, 54, and 108. The common factors of 84 and 108 are 1, 2, 3, 4, 6, and 12.

So, Lekhana can use any of these weights for her bags - 1 kg, 2 kg, 3 kg, 4 kg, 6 kg, or 12 kg. But she wants to use as few bags as possible. Which weight should she choose? If she uses 1-kg bags, she will need 84 bags for the first farm and 108 bags for the second farm - that is a lot! If she uses 12-kg bags, she will need only 7 bags for the first farm and 9 bags for the second farm. That is much fewer bags. So, she should choose the largest common factor, which is 12 kg. So each bag should weigh 12 kilograms.

Now, let us think about a game you might have played in Grade 6 called "Jump Jackpot". In this game, Grumpy places a treasure on a number, and Jumpy chooses a jump size and tries to land on the number with the treasure. Let us consider some examples. Suppose the treasures are placed on numbers 14 and 30. What is the longest jump size that Jumpy can use, starting from 0, so that he lands on both 14 and 30? This is exactly finding the HCF of 14 and 30! The factors of 14 are 1, 2, 7, and 14. The factors of 30 are 1, 2, 3, 5, 6, 10, 15, and 30. The common factors are 1 and 2. So the longest jump size is 2. Similarly, for 7 and 11, the factors of 7 are 1 and 7, and the factors of 11 are 1 and 11. The only common factor is 1. So the longest jump size is 1. For 30 and 50, the common factors are 1, 2, 5, and 10, so the longest jump size is 10. For 28 and 42, the common factors are 1, 2, 7, and 14, so the longest jump size is 14.

Is the longest jump size for these numbers the same as their HCF? Yes, it is! This makes sense because if Jumpy jumps by the HCF each time, he will land on multiples of the HCF. And if the HCF is a factor of both numbers, then both numbers are multiples of the HCF, so he will land on both of them. And since the HCF is the greatest common factor, it is the largest jump size that will work.

Now, so far we have been listing all the factors to find the HCF. But this can become very cumbersome when the numbers have many factors. For example, for 30 and 50, we had to list quite a few factors. And for 28 and 42, it was also a bit tedious. Sometimes, we might also miss some factors, which can lead to errors. So, can we simplify this process? Can we make it more reliable?

It turns out that using prime factorisation can simplify the process of finding the HCF. But before we learn that, let us revisit what we know about primes and prime factorisation.

Do you remember what a prime number is? A prime is a number greater than 1 that has only 1 and the number itself as its factors. For example, 2, 3, 5, 7, 11, 13, and so on are prime numbers. Last year, you learned about the Sieve of Eratosthenes, which is a method to list all the prime numbers up to a certain limit.

Now, what is prime factorisation? Any composite number can be written as a product of prime numbers. We keep rewriting composite factors until only primes are left. This is called the prime factorisation of the given number. For example, let us find the prime factorisation of 90. We can write 90 as 9 times 10. Now, 9 is 3 times 3, and 10 is 2 times 5. So, 90 equals 3 times 3 times 2 times 5. We can reorder this as 2 times 3 times 3 times 5. So the prime factorisation of 90 is 2 × 3 × 3 × 5, or 2 × 3² × 5.

The number 90 could also have been factorised as 3 times 30, or 2 times 45, or in several other different ways. Will these all lead to the same prime factors? Remarkably, yes! The resulting prime factors are always the same, with perhaps only a change in their order. This is a very important property called the Fundamental Theorem of Arithmetic, which says that every composite number can be written as a product of prime factors, and this representation is unique, except for the order of the factors. For example, if we factorise 90 as 3 times 30, and then factorise 30 as 3 times 10, and then factorise 10 as 2 times 5, we get 90 equals 3 times 3 times 2 times 5, which is the same as before. If we start with 2 times 45, and factorise 45 as 3 times 15, and then 15 as 3 times 5, we get 90 equals 2 times 3 times 3 times 5, again the same. So, no matter how we factorise, we always get the same prime factors.

Note that the prime factorisation of a prime number is just the prime number itself. For example, the prime factorisation of 7 is simply 7.

Now, let us learn the procedure for prime factorisation, which is also called the division method. Let me show you how it works with an example. Suppose we want to find the prime factorisation of 105. We start by dividing 105 by the smallest prime, which is 2. But 105 is not divisible by 2, so we try the next prime, which is 3. 105 divided by 3 is 35. So we write 3 on the left and 35 below. Now, 35 is divisible by 5, so we write 5 on the left and 7 below. Now 7 is a prime, so we stop. The prime factors are 3, 5, and 7. So 105 equals 3 times 5 times 7.

Similarly, for 30, we start by dividing by 2. 30 divided by 2 is 15. So we write 2 on the left and 15 below. Now 15 is divisible by 3, so we write 3 on the left and 5 below. Now 5 is a prime, so we stop. The prime factors are 2, 3, and 5. So 30 equals 2 times 3 times 5.

This method is very systematic and helps us find the prime factorisation without missing any factors. Let us try another example. Find the prime factorisation of 1200. We start by dividing by 2. 1200 divided by 2 is 600. Divide by 2 again: 600 divided by 2 is 300. Divide by 2 again: 300 divided by 2 is 150. Divide by 2 again: 150 divided by 2 is 75. Now 75 is not divisible by 2, so we try 3. 75 divided by 3 is 25. Now 25 is divisible by 5. 25 divided by 5 is 5. And 5 divided by 5 is 1. So the prime factors are 2, 2, 2, 2, 3, 5, and 5. So 1200 equals 2 × 2 × 2 × 2 × 3 × 5 × 5, or 2⁴ × 3 × 5².

If we had used the earlier method of randomly factorising, our calculation might have been more complicated. For example, we could have written 1200 as 40 times 30, and then factorised each part. But the division method is much cleaner and more reliable.

Now, why is prime factorisation useful? It helps us find all the factors of a number. Consider the number 840. Its prime factorisation is 2 × 2 × 2 × 3 × 5 × 7. Now, is 2 × 2 × 7, which is 28, a factor of 840? Let us check. If 28 is a factor of 840, then there must be another number that, when multiplied by 28, gives 840. Let us reorder the prime factors of 840. We can write 840 as (2 × 2 × 7) × (2 × 3 × 5). That is 28 × 30. So yes, 28 is a factor of 840, and it should be multiplied by 30 to get 840.

Similarly, is 2 × 7, which is 14, a factor of 840? Let us see. We can write 840 as (2 × 7) × (2 × 2 × 3 × 5). That is 14 × 60. So yes, 14 is also a factor of 840.

Is 2 × 2 × 2, which is 8, a factor of 840? Yes, because we can write 840 as 8 × 105.

Is 3 × 3 × 3, which is 27, a factor of 840? Let us check. 840 divided by 27 is not a whole number. In fact, 840 has only one factor of 3 in its prime factorisation, so it cannot have three 3s as a factor. So 27 is not a factor of 840.

So, we can use the prime factorisation to determine whether one number is a factor of another. And we can also list all the factors of a number by taking all possible combinations of the prime factors.

For example, let us find the factors of 225 using its prime factorisation. First, we find the prime factorisation of 225. Using the division method: 225 divided by 5 is 45. 45 divided by 5 is 9. 9 divided by 3 is 3. 3 divided by 3 is 1. So the prime factorisation of 225 is 5 × 5 × 3 × 3, or 3² × 5².

Now, to find all the factors, we need to consider all possible subparts of this factorisation. First, we have the single prime factors: 3 and 5. Then, we have combinations of two prime factors: 3 × 3 = 9, 5 × 5 = 25, and 3 × 5 = 15. Then, we have combinations of three prime factors: 3 × 3 × 5 = 45, and 3 × 5 × 5 = 75. Then, we have all four prime factors: 3 × 3 × 5 × 5 = 225. And of course, 1 is always a factor. So the factors of 225 are 1, 3, 5, 9, 15, 25, 45, 75, and 225. Let us check if we missed anything. Yes, these are all the factors of 225.

Now, let us practice finding factors using prime factorisation. Find the factors of 90. The prime factorisation of 90 is 2 × 3 × 3 × 5. So the factors are: 1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45, and 90. Can you verify this?

Find the factors of 105. The prime factorisation of 105 is 3 × 5 × 7. So the factors are: 1, 3, 5, 7, 15, 21, 35, and 105.

Find the factors of 132. The prime factorisation of 132 is 2 × 2 × 3 × 11. So the factors are: 1, 2, 3, 4, 6, 11, 12, 22, 33, 44, 66, and 132.

Find the factors of 360. The prime factorisation of 360 is 2 × 2 × 2 × 3 × 3 × 5. This number has 24 factors. Can you list them all?

Find the factors of 840. The prime factorisation of 840 is 2 × 2 × 2 × 3 × 5 × 7. This number has 32 factors.

Now, after observing a few prime factorisations, a student named Anshu claims that "the larger a number is, the longer its prime factorisation will be." What do you think of this claim? Let us check with some examples. Consider the number 96. Its prime factorisation is 2 × 2 × 2 × 2 × 2 × 3, which has six prime factors. Now consider the number 121. Its prime factorisation is 11 × 11, which has only two prime factors. But 121 is larger than 96, yet its prime factorisation is shorter! So Anshu's claim is not true. In mathematics, statements or claims made without proof or verification are called conjectures. Anshu's claim is a conjecture. We disproved this conjecture by finding a counterexample, which is an example where the conjecture is false. This is an important lesson: we should always verify our conjectures with multiple examples before accepting them as true.

Now, let us learn how to find the HCF of two numbers using prime factorisation. This is a very efficient method, especially for large numbers.

Consider Example 1: Find the common factors and the HCF of 45 and 75.

First, we find the prime factorisation of both numbers. 45 equals 3 × 3 × 5. 75 equals 3 × 5 × 5.

Now, the common factors should be subparts that appear in both factorisations. Let us list them. The number 3 appears in both, so 3 is a common factor. The number 5 appears in both, so 5 is a common factor. The number 3 × 5, which is 15, also appears in both, so 15 is a common factor. And of course, 1 is always a common factor. So the common factors are 1, 3, 5, and 15. The highest among them is 15. So the HCF of 45 and 75 is 15.

Now, let us look at Example 2: Find the common factors and the HCF of 112 and 84.

First, we find the prime factorisation. 112 equals 2 × 2 × 2 × 2 × 7, or 2⁴ × 7. 84 equals 2 × 2 × 3 × 7, or 2² × 3 × 7.

Now, let us find the common subparts. Both have the prime factor 2. 112 has four 2s, and 84 has two 2s. So the common part can have at most two 2s. Both have the prime factor 7. So the common part can have one 7. So the common factors are: 2, 2 × 2 = 4, 7, 2 × 7 = 14, 2 × 2 × 7 = 28. And of course, 1. So the common factors are 1, 2, 4, 7, 14, and 28. The highest among them is 28. So the HCF of 112 and 84 is 28.

Now, Example 3: Find the common factors and the HCF of 96 and 275.

First, prime factorisation: 96 equals 2 × 2 × 2 × 2 × 2 × 3, or 2⁵ × 3. 275 equals 5 × 5 × 11, or 5² × 11.

Now, do these two factorisations have any common primes? Let us see. 96 has primes 2 and 3. 275 has primes 5 and 11. There is no prime that appears in both factorisations. So there is no common subpart other than 1. So the only common factor is 1. And 1 is also the HCF. So the HCF of 96 and 275 is 1.

Now, let us think about how we can directly find the HCF without listing all the factors. This is the key insight: to find the HCF, we identify the common primes and find the minimum number of times each of them appears in the factorisations of the given numbers.

Consider Example 4: Find the HCF of 30 and 72.

30 equals 2 × 3 × 5. 72 equals 2 × 2 × 2 × 3 × 3, or 2³ × 3².

Now, the common primes are 2 and 3. For the prime 2, 30 has one 2, and 72 has three 2s. The minimum number is one. So we include one 2 in the HCF. For the prime 3, 30 has one 3, and 72 has two 3s. The minimum number is one. So we include one 3 in the HCF. So the HCF is 2 × 3 = 6.

Now, Example 5: Find the HCF of 225 and 750.

225 equals 3 × 3 × 5 × 5, or 3² × 5². 750 equals 2 × 3 × 5 × 5 × 5, or 2 × 3 × 5³.

The common primes are 3 and 5. For the prime 3, 225 has two 3s, and 750 has one 3. The minimum is one. So we include one 3. For the prime 5, 225 has two 5s, and 750 has three 5s. The minimum is two. So we include two 5s. So the HCF is 3 × 5 × 5 = 75.

This method works for more than two numbers as well. We just need to find the minimum number of occurrences of each common prime across all the numbers.

Now, let us try some practice problems. Find the HCF of 24 and 180. 24 equals 2³ × 3. 180 equals 2² × 3² × 5. The common primes are 2 and 3. Minimum of 2 is 2, minimum of 3 is 1. So HCF is 2² × 3 = 4 × 3 = 12.

Find the HCF of 42, 75, and 24. 42 equals 2 × 3 × 7. 75 equals 3 × 5². 24 equals 2³ × 3. The common prime is only 3. Minimum occurrences of 3 is 1. So HCF is 3.

Find the HCF of 240 and 378. 240 equals 2⁴ × 3 × 5. 378 equals 2 × 3³ × 7. Common primes are 2 and 3. Minimum of 2 is 1, minimum of 3 is 1. So HCF is 2 × 3 = 6.

Find the HCF of 400 and 2500. 400 equals 2⁴ × 5². 2500 equals 2² × 5⁴. Common primes are 2 and 5. Minimum of 2 is 2, minimum of 5 is 2. So HCF is 2² × 5² = 4 × 25 = 100.

Find the HCF of 300 and 800. 300 equals 2² × 3 × 5². 800 equals 2⁵ × 5². Common primes are 2 and 5. Minimum of 2 is 2, minimum of 5 is 2. So HCF is 2² × 5² = 4 × 25 = 100.

Now, before we move on, let me ask you a question. Consider the numbers 72 and 144. Suppose they are factorised into composite numbers as 72 equals 6 times 12, and 144 equals 8 times 18. Seeing this, can we say that these two numbers have no common factor other than 1? Why not? No, we cannot say that, because 6 and 12 can be factorised further. 6 equals 2 times 3, and 12 equals 2 times 2 times 3. So 72 actually equals 2 times 3 times 2 times 2 times 3, which is 2³ × 3². Similarly, 8 equals 2³, and 18 equals 2 times 3², so 144 equals 2⁴ × 3². The common factors are 2³ and 3², so the HCF is 2³ × 3² = 8 × 9 = 72. So we must always factorise completely into prime factors to find the HCF correctly.

Now, let us move on to the second major concept of this chapter: the Lowest Common Multiple, or LCM.

Let us start with another story. Two friends, Anshu and Guna, make torans, which are decorative strings made from strips of cloth. Multiple strips are placed one next to another to make a toran. Anshu uses strips of length 6 cm, and Guna uses strips of 8 cm length. If both have to make torans of the same length, what is the smallest possible length the torans could be?

Anshu uses cloth strips of 6 cm each. So any toran he makes will be a multiple of 6. So the length of the toran could be 6 cm, 12 cm, 18 cm, 24 cm, 30 cm, 36 cm, 42 cm, 48 cm, 54 cm, and so on.

Similarly, any toran Guna makes should be a multiple of 8. So the length of the toran he makes could be 8 cm, 16 cm, 24 cm, 32 cm, 40 cm, 48 cm, 56 cm, 64 cm, 72 cm, and so on.

From this, we can see that if both have to make torans of the same length, the length of the toran should be a common multiple of 6 and 8. From the two lists, we can see that 24 cm and 48 cm are two of the common multiples of 6 and 8. So 24 cm and 48 cm are lengths of toran that both Anshu and Guna can make.

But 24 cm is the smallest among all the common multiples. So 24 cm is the length of the shortest toran that both can stitch. This is the lowest common multiple.

Now, what about the largest common multiple? Does such a number exist? No, there is no largest common multiple, because multiples go on forever. We can always find a larger common multiple by multiplying the LCM by any number.

Now, let us consider another example. A sweet shop gives out free gajak to school children on Mondays. Gajak is a sweet made from sesame seeds, jaggery, and ghee. Today is a Monday, and a girl named Kabamai enjoyed eating the gajak. But she visits the sweet shop once every 10 days. When is the next time she would be able to get free gajak from the sweet shop?

Since the shop gives free sweets every Monday, it will give free sweets again after 7 days, 14 days, 21 days, 28 days, 35 days, 42 days, 49 days, 56 days, 63 days, 70 days, and so on. These are multiples of 7.

Kabamai will arrive at the sweet shop again after 10 days, 20 days, 30 days, 40 days, 50 days, 60 days, 70 days, and so on. These are multiples of 10.

When will Kabamai eat free sweets again? It will happen on days that are common to both sequences. Looking at the two lists, we can see that the first common day is 70 days. So after 70 days, Kabamai will be able to get free gajak again.

Notice that here too, 70 is the lowest among all the common multiples of 7 and 10. So this is another example of finding the lowest common multiple.

The Lowest Common Multiple, or LCM, of two or more given numbers is the lowest, or smallest, or least, of their common multiples.

Do you remember the "Idli-Vada" game from Grade 6? Two numbers are chosen, and whenever players come to their multiples, "idli" or "vada" should be called out depending on whose multiple the number is. If the number happens to be a common multiple, then "idli-vada" should be called out. In each problem below, the two numbers corresponding to "idli" and "vada" are given. Find the first number for which "idli-vada" will be called out.

For 4 and 6, the multiples of 4 are 4, 8, 12, 16, 20, 24, and so on. The multiples of 6 are 6, 12, 18, 24, 30, and so on. The first common multiple is 12. So the answer is 12, which is the LCM of 4 and 6.

For 7 and 11, the multiples of 7 are 7, 14, 21, 28, 35, 42, 49, 56, 63, 70, 77, and so on. The multiples of 11 are 11, 22, 33, 44, 55, 66, 77, and so on. The first common multiple is 77. So the answer is 77, which is the LCM of 7 and 11.

For 14 and 30, the multiples of 14 are 14, 28, 42, 56, 70, 84, 98, 112, 126, 140, and so on. The multiples of 30 are 30, 60, 90, 120, 150, and so on. The first common multiple is 210. So the answer is 210, which is the LCM of 14 and 30.

For 15 and 55, the multiples of 15 are 15, 30, 45, 60, 75, 90, 105, 120, 135, 150, 165, 180, 195, 210, 225, 240, 255, 270, 285, 300, 315, 330, 345, 360, 375, 390, 405, 420, 435, 450, 465, 480, 495, 510, 525, 540, 555, 570, 585, 600, 615, 630, 645, 660, 675, 690, 705, 720, 735, 750, 765, 780, 795, 810, 825, and so on. The multiples of 55 are 55, 110, 165, 220, 275, 330, 385, 440, 495, 550, 605, 660, 715, 770, 825, and so on. The first common multiple is 165. So the answer is 165, which is the LCM of 15 and 55.

Is the answer always the LCM of the two numbers? Yes, it is! The first number on which both players would say "idli-vada" is exactly the LCM of the two numbers.

As in the case of the HCF, the process of finding the LCM by listing down the multiples may get tedious for larger numbers, as we saw for 14 and 30, and 15 and 55. So, we need a more efficient method, and that is where prime factorisation comes in handy again.

How do we find the LCM of two numbers using their prime factors? We have seen that every factor of a number is formed by taking a subpart of its prime factorisation. We used this fact to come up with a method to find the HCF. In a similar manner, we can come up with a method to find the LCM.

We begin by comparing the prime factorisations of a number and a multiple of that number. For example, let us take 36 and its multiple 648, which equals 36 times 18.

We get 36 equals 2 × 2 × 3 × 3. And 648 equals 36 × 18, which is (2 × 2 × 3 × 3) × (2 × 3 × 3). So 648 equals 2 × 2 × 2 × 3 × 3 × 3 × 3.

What do we observe? We can see that the prime factors of the multiple contain the prime factors of the number along with some more prime factors. Will this happen with every multiple? Yes, it will! Because if a number is a multiple of another number, then it must contain all the prime factors of the original number, plus some additional prime factors.

Can this be used to find the LCM? Yes, it can! The LCM must contain all the prime factors that appear in either number, and for each prime, it must contain the maximum number of times that prime appears in either number.

Now, let us look at Example 6: Find the LCM of 14 and 35.

First, we find the prime factorisations. 14 equals 2 × 7. 35 equals 5 × 7.

Now, what are the prime factors that appear in either number? They are 2, 5, and 7.

Now, how many times does each prime appear? For 2, it appears once in 14 and zero times in 35. The maximum is once. For 5, it appears zero times in 14 and once in 35. The maximum is once. For 7, it appears once in both 14 and 35. The maximum is once.

So the LCM should be 2 × 5 × 7 = 70.

Let us verify. Is 70 a multiple of 14? Yes, 14 × 5 = 70. Is 70 a multiple of 35? Yes, 35 × 2 = 70. Is there any smaller common multiple? Let us check. 14 × 2 = 28, but 28 is not a multiple of 35. 14 × 3 = 42, not a multiple of 35. 14 × 4 = 56, not a multiple of 35. 14 × 5 = 70, which works. So 70 is indeed the smallest common multiple, which is the LCM.

Now, Example 7: Find the LCM of 96 and 360.

First, prime factorisation: 96 equals 2 × 2 × 2 × 2 × 2 × 3, or 2⁵ × 3. 360 equals 2 × 2 × 2 × 3 × 3 × 5, or 2³ × 3² × 5.

Now, the prime factors appearing in either number are 2, 3, and 5.

For the prime 2, 96 has five 2s, and 360 has three 2s. The maximum is five. So we need five 2s in the LCM.

For the prime 3, 96 has one 3, and 360 has two 3s. The maximum is two. So we need two 3s in the LCM.

For the prime 5, 96 has zero 5s, and 360 has one 5. The maximum is one. So we need one 5 in the LCM.

Thus, the LCM is 2 × 2 × 2 × 2 × 2 × 3 × 3 × 5, which equals 32 × 9 × 5, which equals 1440.

Let us verify. Is 1440 a multiple of 96? 96 × 15 = 1440, yes. Is 1440 a multiple of 360? 360 × 4 = 1440, yes. Is there any smaller common multiple? If we try to remove any prime factor from 1440, it would no longer be a multiple of both numbers. So 1440 is indeed the LCM.

So, to find the LCM of two numbers, we identify all the prime factors and find the maximum number of times each of them occurs in either of the factorisations. This process can be extended to find the LCM of more than two numbers as well.

Now, let us practice finding the LCM using prime factorisation.

Find the LCM of 30 and 72. 30 equals 2 × 3 × 5. 72 equals 2³ × 3². The prime factors are 2, 3, and 5. Maximum of 2 is 3, maximum of 3 is 2, maximum of 5 is 1. So LCM is 2³ × 3² × 5 = 8 × 9 × 5 = 360.

Find the LCM of 36 and 54. 36 equals 2² × 3². 54 equals 2 × 3³. Prime factors are 2 and 3. Maximum of 2 is 2, maximum of 3 is 3. So LCM is 2² × 3³ = 4 × 27 = 108.

Find the LCM of 105, 195, and 65. 105 equals 3 × 5 × 7. 195 equals 3 × 5 × 13. 65 equals 5 × 13. Prime factors are 3, 5, 7, and 13. Maximum of 3 is 1, maximum of 5 is 1, maximum of 7 is 1, maximum of 13 is 1. So LCM is 3 × 5 × 7 × 13 = 1365.

Find the LCM of 222 and 370. 222 equals 2 × 3 × 37. 370 equals 2 × 5 × 37. Prime factors are 2, 3, 5, and 37. Maximum of 2 is 1, maximum of 3 is 1, maximum of 5 is 1, maximum of 37 is 1. So LCM is 2 × 3 × 5 × 37 = 1110.

Now, let us move on to the third section of this chapter: Patterns, Properties, and a Pretty Procedure!

We have learned how to find HCF and LCM using prime factorisation. Now, let us explore some interesting patterns and properties.

First, let us look at some special cases. The HCF of 6 and 18 is 6, which is one of the two given numbers. Can we find more such number pairs where the HCF is one of the two numbers?

Let us think about this. When does the HCF equal one of the numbers? This happens when one number is a factor of the other. For example, 6 is a factor of 18, so the HCF of 6 and 18 is 6. Similarly, the HCF of 5 and 15 is 5. The HCF of 7 and 28 is 7. The HCF of 12 and 48 is 12. In each case, one number is a factor of the other, and the smaller number is the HCF.

How can we describe such pairs of numbers using algebra? If n is a number, then any multiple of n can be written as a positive integer multiplied by n. For example, if we take n and 5n, then 5n is a multiple of n, and n is a factor of 5n. The HCF of n and 5n is n.

So, for any number m, the other number could be any multiple of m, like 2m, 3m, 4m, and so on. Similarly, if 7k is a number, the other number could be any multiple of 7k, like 14k, 21k, 28k, and so on.

Now, let us explore some more patterns. Consider two consecutive even numbers, like 14 and 16, or 28 and 30. What is their HCF? The HCF of 14 and 16 is 2. The HCF of 28 and 30 is 2. It seems like the HCF of any two consecutive even numbers is always 2. Is this true? Let us think about why. Two consecutive even numbers differ by 2. They are both divisible by 2. But they cannot have any larger common factor, because if they had a common factor greater than 2, then that factor would also divide their difference, which is 2. The only factors of 2 are 1 and 2. Since both numbers are even, 2 is a common factor. So the HCF is 2.

Now, consider two consecutive odd numbers, like 13 and 15, or 27 and 29. What is their HCF? The HCF of 13 and 15 is 1. The HCF of 27 and 29 is 1. It seems like the HCF of any two consecutive odd numbers is always 1. Why? Because consecutive odd numbers differ by 2. If they had a common factor greater than 1, that factor would also divide their difference, which is 2. But the only common factor they can have is 1, because odd numbers are not divisible by 2. So the HCF is 1.

Now, consider two even numbers, not necessarily consecutive, like 12 and 20, or 18 and 30. What can we say about their HCF? Well, since both are even, they are both divisible by 2. So 2 is always a common factor. But the HCF could be more than 2. For example, the HCF of 12 and 20 is 4, and the HCF of 18 and 30 is 6. So we cannot make a general statement about the HCF of any two even numbers except that it is at least 2.

Now, consider two consecutive numbers, like 15 and 16, or 29 and 30. What is their HCF? The HCF of 15 and 16 is 1. The HCF of 29 and 30 is 1. It seems like the HCF of any two consecutive numbers is always 1. Why? Because consecutive numbers differ by 1. If they had a common factor greater than 1, that factor would also divide their difference, which is 1. But the only factor of 1 is 1 itself. So the HCF is 1.

Now, consider two co-prime numbers. Do you remember what co-prime numbers are? Two numbers are co-prime if their HCF is 1. For example, 4 and 9 are co-prime because their HCF is 1. So by definition, the HCF of two co-prime numbers is 1.

So we have made some general statements: The HCF of two consecutive even numbers is 2. The HCF of two consecutive odd numbers is 1. The HCF of two consecutive numbers is 1. The HCF of two co-prime numbers is 1. These are examples of generalisation - making a statement that describes a pattern or property that holds in all possible cases.

Now, let us look at similar patterns for LCM. The LCM of 3 and 24 is 24, which is one of the two given numbers. Find more such number pairs where the LCM is one of the two numbers.

This happens when one number is a multiple of the other. For example, the LCM of 5 and 15 is 15. The LCM of 7 and 28 is 28. The LCM of 12 and 48 is 48. In each case, the larger number is a multiple of the smaller one, so the larger number is the LCM.

Using algebra, if n is a number, then any multiple of n can be written as n times some positive integer. So if we take n and kn, the LCM is kn, which is the larger number.

Now, let us make general statements about the LCM for different pairs of numbers.

First, consider two multiples of 3. For example, 12 and 18 are both multiples of 3. The LCM of 12 and 18 is 36. Is 36 also a multiple of 3? Yes, it is. In fact, the LCM of any two multiples of 3 will also be a multiple of 3. More generally, if two numbers are both multiples of some number n, then their LCM will also be a multiple of n.

Second, consider two consecutive even numbers, like 14 and 16. The LCM of 14 and 16 is 112. Is 112 a multiple of both 14 and 16? Yes, it is. But is there a simpler way to find it? Not really, we need to calculate it.

Third, consider two consecutive numbers, like 15 and 16. The LCM of 15 and 16 is 240. Since consecutive numbers are co-prime, their LCM is simply their product. In general, the LCM of two consecutive numbers is their product.

Fourth, consider two co-prime numbers, like 4 and 9. The LCM of 4 and 9 is 36, which is 4 times 9. In general, the LCM of two co-prime numbers is their product.

Now, let us explore what happens to the HCF of two numbers if both numbers are doubled. Take some pairs of numbers and explore. For example, consider 12 and 18. Their HCF is 6. Now, if we double both numbers, we get 24 and 36. The HCF of 24 and 36 is 12. Notice that 12 is double 6. Let us try another example. Consider 10 and 15. Their HCF is 5. Double both to get 20 and 30. The HCF of 20 and 30 is 10, which is double 5. It seems like if both numbers are doubled, the HCF also doubles. Why does this happen? Because if we double both numbers, we are essentially multiplying each by 2. This adds an extra factor of 2 to each number's prime factorisation. Since this factor of 2 is common to both numbers, it gets included in the HCF as well. So the HCF also gets multiplied by 2. In general, if both numbers are multiplied by some factor k, the HCF is also multiplied by k, provided that k is a factor of both numbers. Actually, more precisely, if both numbers are multiplied by the same number, their HCF is also multiplied by that number.

Let us verify with an example. Consider the numbers 270 and 50. 270 equals 2 × 3 × 3 × 3 × 5. 50 equals 2 × 5 × 5. The HCF is 2 × 5 = 10. Now, let us double these numbers to get 540 and 100. 540 equals 2 × 2 × 3 × 3 × 3 × 5. 100 equals 2 × 2 × 5 × 5. The HCF is 2 × 2 × 5 = 20, which is indeed double 10.

Now, consider two multiples of the same number. For example, consider 14 × 6 and 14 × 9. What is their HCF? Clearly, 14 is a common factor. But is it the highest? Let us calculate. 14 × 6 equals 2 × 7 × 2 × 3, which is 2² × 3 × 7. 14 × 9 equals 2 × 7 × 3 × 3, which is 2 × 3² × 7. The common factors are 2 and 7. The minimum number of 2s is 1, and we have one 7. So the HCF is 2 × 7 = 14. But wait, we also have a common factor of 3 in both? No, 14 × 6 has one 3, and 14 × 9 has two 3s, so the minimum is one 3. But 3 is not a common prime factor of the original numbers? Actually, let us look more carefully. 14 × 6 = 84, and 14 × 9 = 126. The HCF of 84 and 126 is 42. And 42 equals 14 × 3. So in this case, the HCF is not just 14, but 14 times the HCF of 6 and 9, which is 3. So the HCF is 14 × 3 = 42.

So, for two numbers that are both multiples of the same number, say n × a and n × b, the HCF is n times the HCF of a and b. In other words, HCF(n × a, n × b) = n × HCF(a, b).

Let us verify with some more examples. Consider 18 × 10 and 18 × 15. 18 × 10 = 180, 18 × 15 = 270. The HCF of 10 and 15 is 5. So the HCF should be 18 × 5 = 90. Let us check: the factors of 180 are 1, 2, 3, 4, 5, 6, 9, 10, 12, 15, 18, 20, 30, 36, 45, 60, 90, 180. The factors of 270 are 1, 2, 3, 5, 6, 9, 10, 15, 18, 27, 30, 45, 54, 90, 135, 270. The common factors are 1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45, 90. The highest is 90. Yes, it is 90.

Consider 10 × 38 and 10 × 21. 10 × 38 = 380, 10 × 21 = 210. The HCF of 38 and 21 is 1 (since 38 = 2 × 19, and 21 = 3 × 7, no common primes). So the HCF should be 10 × 1 = 10. Let us check: the factors of 380 include 10, and the factors of 210 include 10. Is there any larger common factor? 380 = 2² × 5 × 19, 210 = 2 × 3 × 5 × 7. The common primes are 2 and 5. The minimum is one each, so HCF is 2 × 5 = 10. Yes, it is 10.

Consider 5 × 13 and 5 × 20. 5 × 13 = 65, 5 × 20 = 100. The HCF of 13 and 20 is 1. So the HCF should be 5 × 1 = 5. Let us check: 65 = 5 × 13, 100 = 2² × 5². The common prime is 5, so HCF is 5. Yes.

Consider 12 × 16 and 12 × 20. 12 × 16 = 192, 12 × 20 = 240. The HCF of 16 and 20 is 4. So the HCF should be 12 × 4 = 48. Let us check: 192 = 2⁶ × 3, 240 = 2⁴ × 3 × 5. The common primes are 2 and 3. Minimum of 2 is 4, minimum of 3 is 1. So HCF is 2⁴ × 3 = 16 × 3 = 48. Yes.

In which of these cases is the HCF the same as the common multiplier? In the second example, where we had 10 × 38 and 10 × 21, the HCF was 10, which is the same as the common multiplier. This happened because the HCF of 38 and 21 was 1. So when the two numbers being multiplied have no common factor, the HCF is just the common multiplier itself.

Now, let us learn an efficient procedure for finding both HCF and LCM at the same time. This is sometimes called the "ladder method" or "division method".

Let me show you how it works. Suppose we want to find the HCF and LCM of 84 and 180.

First, we write the two numbers side by side, separated by a comma. Then we try to divide both by a common prime factor. We start with 2, since both numbers are even.

2 goes into 84, and 2 goes into 180. So we divide both by 2. We get 42 and 90. We write 2 on the left.

Now, 42 and 90 are still both even, so we can divide by 2 again. 42 divided by 2 is 21, and 90 divided by 2 is 45. We write another 2 on the left. Now we have 21 and 45.

Now, 21 and 45 are not both even anymore. Let us try the next prime, which is 3. 21 divided by 3 is 7, and 45 divided by 3 is 15. We write 3 on the left. Now we have 7 and 15.

Now, 7 and 15 have no common prime factors (7 is prime, 15 is 3 times 5). So we stop here.

Now, to find the HCF, we multiply all the prime factors we wrote on the left. So HCF = 2 × 2 × 3 = 12.

To find the LCM, we multiply all the prime factors on the left by the numbers in the last row. So LCM = 2 × 2 × 3 × 7 × 15 = 2 × 2 × 3 × 7 × 15. Let us calculate: 2 × 2 = 4, 4 × 3 = 12, 12 × 7 = 84, 84 × 15 = 1260. So the LCM is 1260.

Let us verify. Is 1260 a multiple of 84? 84 × 15 = 1260, yes. Is 1260 a multiple of 180? 180 × 7 = 1260, yes. And is 12 the HCF? 84 divided by 12 is 7, and 180 divided by 12 is 15. And 7 and 15 have no common factors. So yes, HCF is 12 and LCM is 1260.

Why does this work? Because when we divide both numbers by a common prime factor, we are essentially factoring out the common part. The product of all the common prime factors gives us the HCF. And for the LCM, we need to include all the prime factors from both numbers. The numbers in the last row, 7 and 15, represent the parts that are not common. So when we multiply the common factors (2, 2, 3) by the non-common parts (7 and 15), we get a number that contains all prime factors from both original numbers, which is the LCM.

Let us try this method with another example. Find the HCF and LCM of 300 and 150.

We write 300 and 150. Both are divisible by 2. 300 ÷ 2 = 150, 150 ÷ 2 = 75. So we write 2 on the left. Now we have 150 and 75.

Both are divisible by 3. 150 ÷ 3 = 50, 75 ÷ 3 = 25. So we write 3 on the left. Now we have 50 and 25.

Both are divisible by 5. 50 ÷ 5 = 10, 25 ÷ 5 = 5. So we write 5 on the left. Now we have 10 and 5.

Both are divisible by 5 again. 10 ÷ 5 = 2, 5 ÷ 5 = 1. So we write 5 on the left. Now we have 2 and 1.

Now we stop, because 1 has no more factors.

The HCF is the product of all the factors on the left: 2 × 3 × 5 × 5 = 150.

The LCM is the product of all the factors on the left multiplied by the numbers in the last row: 2 × 3 × 5 × 5 × 2 × 1 = 2 × 3 × 5 × 5 × 2 = 300.

So HCF is 150 and LCM is 300. Notice that in this case, 300 is exactly twice 150. And indeed, 150 is a factor of 300, so the HCF is the smaller number and the LCM is the larger number.

Let us try one more. Find the HCF and LCM of 630 and 770.

We write 630 and 770. Both are divisible by 2. 630 ÷ 2 = 315, 770 ÷ 2 = 385. Write 2 on the left. Now we have 315 and 385.

Both are divisible by 5. 315 ÷ 5 = 63, 385 ÷ 5 = 77. Write 5 on the left. Now we have 63 and 77.

Both are divisible by 7. 63 ÷ 7 = 9, 77 ÷ 7 = 11. Write 7 on the left. Now we have 9 and 11.

Now, 9 and 11 have no common factors, so we stop.

The HCF is 2 × 5 × 7 = 70.

The LCM is 2 × 5 × 7 × 9 × 11 = 2 × 5 × 7 × 99 = 70 × 99 = 6930.

So HCF is 70 and LCM is 6930.

Now, there is also a shortcut we can use. Instead of dividing by one prime at a time, we can divide by any common factor we can find. For example, for 300 and 150, we could have divided both by 50 directly. 300 ÷ 50 = 6, 150 ÷ 50 = 3. Then we could divide by 3: 6 ÷ 3 = 2, 3 ÷ 3 = 1. So the HCF would be 50 × 3 = 150, and the LCM would be 50 × 3 × 2 × 1 = 300. This is the same as before, but we did it in fewer steps.

Similarly, for 630 and 770, we could divide both by 10 first: 630 ÷ 10 = 63, 770 ÷ 10 = 77. Then we can divide both by 7: 63 ÷ 7 = 9, 77 ÷ 7 = 11. So the HCF is 10 × 7 = 70, and the LCM is 10 × 7 × 9 × 11 = 6930. This is faster!

So we can use whatever common factor we can identify at each step to make the process quicker.

Now, let us explore an important property that relates the HCF and LCM of two numbers. Which is greater - the LCM of two numbers or their product? Let us think about this.

Consider the numbers 12 and 18. Their product is 12 × 18 = 216. Their LCM is 36. Is the LCM greater than the product? No, 36 is much less than 216. In fact, the LCM is always less than or equal to the product of the two numbers. Why? Because the product itself is a common multiple of the two numbers. The LCM is the smallest common multiple. So the LCM can never be greater than the product, because the product is one of the common multiples. The LCM might be equal to the product if the two numbers are co-prime, because then the product is the smallest common multiple.

Now, there is an interesting relation between the product of two numbers and their HCF and LCM. Let us explore this with an example.

Consider the numbers 105 and 95. Find their LCM.

First, factorise them: 105 equals 3 × 5 × 7. 95 equals 5 × 19.

The LCM is 3 × 5 × 7 × 19.

Now, consider the product: 105 × 95 equals (3 × 5 × 7) × (5 × 19) = 3 × 5 × 5 × 7 × 19.

Is the LCM a factor of the product? Yes, it is. The LCM is 3 × 5 × 7 × 19. If we multiply this by 5, we get 3 × 5 × 5 × 7 × 19, which is exactly the product. So the product equals LCM × 5.

Now, what is 5? It is the HCF of 105 and 95! Let us check. The common prime factor is 5. So HCF is 5. So we have product = LCM × HCF.

Let us verify with more examples. Consider 45 and 105. 45 equals 3² × 5. 105 equals 3 × 5 × 7. LCM is 3² × 5 × 7 = 315. Product is 45 × 105 = 4725. Is 315 a factor of 4725? 4725 ÷ 315 = 15. And what is the HCF? The common primes are 3 and 5. Minimum of 3 is 1, minimum of 5 is 1. So HCF is 3 × 5 = 15. So product = LCM × HCF.

Consider 275 and 352. 275 equals 5² × 11. 352 equals 2⁵ × 11. LCM is 2⁵ × 5² × 11 = 32 × 25 × 11 = 8800. Product is 275 × 352 = 96800. 96800 ÷ 8800 = 11. And HCF? The common prime is 11. So HCF is 11. Yes, product = LCM × HCF.

Consider 222 and 370. 222 equals 2 × 3 × 37. 370 equals 2 × 5 × 37. LCM is 2 × 3 × 5 × 37 = 1110. Product is 222 × 370 = 82140. 82140 ÷ 1110 = 74. And HCF? The common primes are 2 and 37. So HCF is 2 × 37 = 74. Yes!

So we have discovered a very important property: For any two numbers, HCF × LCM = Product of the two numbers.

Why does this happen? Let us think about it using prime factorisation. Suppose we have two numbers A and B. Let their prime factorisations be A = p₁ᵃ¹ × p₂ᵃ² × ... and B = p₁ᵇ¹ × p₂ᵇ² × ... where p₁, p₂, ... are primes. Some primes may appear in both factorisations, and some may appear in only one.

For each prime, the HCF contains the minimum number of occurrences, and the LCM contains the maximum number of occurrences. So if we multiply HCF and LCM, for each prime, we get the minimum plus the maximum, which equals a₁ + b₁, which is exactly the number of times that prime appears in the product A × B.

So, HCF × LCM = A × B. This is a very useful property that can help us find one of these values if we know the other two.

Now, does this property hold for three numbers? Let us check. Consider three numbers: 2, 3, and 5. HCF of 2, 3, and 5 is 1. LCM is 30. Product is 2 × 3 × 5 = 30. So HCF × LCM = 1 × 30 = 30, which equals the product. Let us try another set: 6, 8, and 12. HCF is 2. LCM is 24. Product is 6 × 8 × 12 = 576. HCF × LCM = 2 × 24 = 48, which is not equal to 576. So the property does not hold for three or more numbers. The property HCF × LCM = Product only works for two numbers.

Now, let us solve some interesting problems using what we have learned.

Problem 1: In the two rows below, colours repeat as shown. When will the blue stars meet next? This seems to be referring to a pattern problem, but without the visual, it is difficult to solve. Let me know if you have the image.

Problem 2: Is 5 × 7 × 11 × 11 a multiple of 5 × 7 × 7 × 11 × 2? Let us check. 5 × 7 × 11 × 11 equals 5 × 7 × 11². 5 × 7 × 7 × 11 × 2 equals 5 × 7 × 7 × 11 × 2. For the first number to be a multiple of the second, the second number must be a factor of the first. But the second number has two 7s and a factor of 2, while the first number has only one 7 and no factor of 2. So no, it is not a multiple.

Is 5 × 7 × 11 × 11 a factor of 5 × 7 × 7 × 11 × 2? For the first to be a factor of the second, the second must contain all the prime factors of the first. The first has 5, 7, 11². The second has 5, 7², 11, 2. The second has two 7s, so it has at least one 7. But it has only one 11, while the first has two 11s. So the first is not a factor of the second. So the answer to both parts is no.

Problem 3: Find the HCF and LCM of the following, and state your answers in the form of prime factorisations.

First, 3 × 3 × 5 × 7 × 7 and 12 × 7 × 11. The first number is 3² × 5 × 7². The second number is 12 × 7 × 11 = (2² × 3) × 7 × 11 = 2² × 3 × 7 × 11. For HCF, we take the minimum of each common prime. Common primes are 3 and 7. Minimum of 3 is 1, minimum of 7 is 1. So HCF is 3 × 7 = 21, or in prime factorisation, 3 × 7. For LCM, we take the maximum of each prime. Primes are 2, 3, 5, 7, 11. Maximum of 2 is 2, maximum of 3 is 2, maximum of 5 is 1, maximum of 7 is 2, maximum of 11 is 1. So LCM is 2² × 3² × 5 × 7² × 11.

Second, 45 and 36. 45 equals 3² × 5. 36 equals 2² × 3². Common prime is 3. Minimum of 3 is 2. So HCF is 3² = 9. For LCM, primes are 2, 3, 5. Maximum of 2 is 2, maximum of 3 is 2, maximum of 5 is 1. So LCM is 2² × 3² × 5 = 4 × 9 × 5 = 180.

Problem 4: Find two numbers whose HCF is 1 and LCM is 66. Since HCF is 1, the numbers are co-prime. And their LCM is 66. Since they are co-prime, their product equals their LCM. So the product is 66. We need to find two co-prime numbers whose product is 66. The factors of 66 are 1, 2, 3, 6, 11, 22, 33, 66. We need two numbers that multiply to 66 and have no common factors other than 1. Let us try 6 and 11. They multiply to 66. Do they have any common factors? 6 = 2 × 3, 11 is prime. So their HCF is 1. So the numbers are 6 and 11.

Problem 5: A cowherd took all his