# FACTORS, PRIME NUMBERS, H.C.F. AND L.C.M.

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1 Mathematics Revision Guides Factors, Prime Numbers, H.C.F. and L.C.M. Page 1 of 10 M.K. HOME TUITION Mathematics Revision Guides Level: GCSE Foundation Tier FACTORS, PRIME NUMBERS, H.C.F. AND L.C.M. Version: 4.2 Date:

2 Mathematics Revision Guides Factors, Prime Numbers, H.C.F. and L.C.M. Page 2 of 10 FACTORS AND MULTIPLES. (These terms relate only to positive whole numbers.) A factor of a number is any number that divides into it without remainder. For example, the factors of 18 are 1, 2, 3, 6, 9 and 18. A multiple of a number is that number multiplied by another positive whole number. Therefore the first five multiples of 12 are 12, 24, 36, 48 and 60 the first entries in the 12-times-table. A number is said to be prime if it has no other factors than itself and 1. The first five prime numbers are 2, 3, 5, 7 and 11. (1 is not considered a prime number as it has only one factor). Example 1. Find the factors of 24. The number 24 can be expressed as a product of two numbers in various ways: 1 24 = = 24 3 = = 24 We can stop here, because the next highest factor of 24, namely 6, has already been included. Once we reach a possible factor greater than the square root of the number, the process is finished, as the products would only reappear in reverse order. The factors of 24 are 1, 2, 3, 4, 6, 8, 12 and 24. When the number is larger, some trial and error might be needed using a calculator. Example 2. Find the factors of 192. The square root of 192 is so there is no need to look for factors over = = = = 192 (5 does not divide exactly into 192, giving an answer of 38.4) 6 32 = 192 (7 does not divide exactly into 192, giving an answer of ) 8 24 = 192 (9 does not divide exactly into 192, giving an answer of ) (10 and 11 do not divide into 192) Also, if a number is not a factor, neither is any of its multiples. We found that 5 was not a factor of 192; therefore 10, being a multiple of 5, cannot be a factor either = 192 (13 does not divide into 192) The factors of 192 are 1, 2, 3, 4, 6, 8, 12, 16, 24, 32, 48, 64, 96 and 192.

3 Mathematics Revision Guides Factors, Prime Numbers, H.C.F. and L.C.M. Page 3 of 10 Example 3. Find the factors of 225. The square root of 225 is exactly 15, so we do not need to look for any factors above it. Because 15 is a factor, so are 3 and 5; also, since 225 is odd, we can forget about any even factors = = = 225 (7 does not divide exactly into 225) 9 25 = 225 (11 does not divide exactly into 225) (13 does not divide exactly into 275) = 225 The factors of 225 are 1, 3, 5, 9, 15, 25, 45, 75 and 225. Factor / divisibility tests for numbers (factors up to 12). (Useful in non-calculator exams, but not essential) There are many short cuts to test if one number divides exactly into another. Although they are not part of the syllabus, some of them can be quite useful time-savers, even in calculator exams. All numbers ending in 2, 4, 6, 8 or 0 (even numbers) are divisible by is divisible by 2, but 37 is not. All numbers whose digit sum is a multiple of 3 is divisible by is divisible by 3, since its digits, 5+1, add up to 6; 58 is not since its digits sum to 13,which is not a multiple of 3. All numbers which are still even after halving are divisible by is divisible by 4 since half of it, 34, is still even; 70 is not because 35 isn t. All numbers ending in 0 or 5 are divisible by is divisible by 5; 82 is not. All even numbers whose digit sum is a multiple of 3 are divisible by is even, and its digits add up to 15, a multiple of 3. Sorry there are no quick tests for divisibility by 7! All numbers which are even after halving them twice running are divisible by is a multiple of 8, as halving it to 52 and then to 26 still leaves an even number. 108 is not, as halving it twice running gives 27, an odd number. All numbers whose digit sum is a multiple of 9 is divisible by is divisible by 9, since its digits, 1+8+9, add up to 18, a multiple of 9; 192 is not since its digits sum to 12,which is not a multiple of 9. Notice, that if you take the digit sum of 18 and add its digits, you will end up with 9. All numbers ending in 0 are multiples of 10. Trivial! All numbers whose alternate digit sums are equal or differ by a multiple of 11 are divisible by 11. This is easier than it sounds 154 is divisible by 11 because the sum of the odd-placed digits (1 and 4) equals the even-placed digit, here is not, because the sum of the odd digits (1 and 6) = 7 and the even digit is 5. The difference between odds and evens is 2. All numbers divisible by both 3 and 4 are divisible by can be halved to give the even value of 66, and its digits add up to 12, a multiple of 3.

4 Mathematics Revision Guides Factors, Prime Numbers, H.C.F. and L.C.M. Page 4 of 10 Testing for prime numbers. To determine if a number is prime or not, we need to try dividing it by every prime number up to its square root. If none of those primes divide into it, then the number is a prime. If there is one counterexample, then the number is not prime. Example (4). Which, if any, of the following numbers are prime? 46, 59, 87, 101, 155, 289? 46 is not a prime because it is even. The square root of 59 lies between 7 and 8, so we need to test if it is divisible by 2, 3, 5 or 7. It is odd (not a multiple of 2), its digits add up to 14 (so it s not a multiple of 3), it does not end in 5 or 0 (indivisible by 5), and it leaves a remainder after dividing by 7. Therefore 59 is prime. The digits of 87 sum to 15, a multiple of 3, so 87 is a multiple of 3 and thus not prime. 101 has a square root less than 11, it is odd, its digits add up to 2 (indivisible by 3), it doesn t end in 5 or 0, and it leaves a remainder after dividing by is thus prime. 155 ends in 5, and is therefore divisible by 5 and hence not prime. 289 has a square root of exactly 17, therefore it is not prime.

5 Mathematics Revision Guides Factors, Prime Numbers, H.C.F. and L.C.M. Page 5 of 10 The Sieve of Eratosthenes. This method of finding prime numbers was demonstrated over two thousand years ago, and we shall use it in the next example. Example (5): Use the Sieve to find all the prime numbers less than 100. The first grid shows all the numbers from 1 to 100. The first prime number is 2, so we keep it, but eliminate, or sieve out all the other even numbers, as they all have a factor of 2 as well as 1 and the number itself. Next, we continue in the same vein by getting rid of all the multiples of 3 excluding 3 itself. We do the same with the multiples of 5 and 7, and finally remove 1 from the grid as it is not technically a prime. Because the square root of 100 is 10, there is no need to test for multiples of 11 and over, as any such numbers less than 101 would have been thrown out in earlier stages. For example, 77 = 7 would have been removed at the multiples of 7 stage.

6 Mathematics Revision Guides Factors, Prime Numbers, H.C.F. and L.C.M. Page 6 of 10 Prime Factors. Because a number can possibly be shown as a product of various factors, such as 24 = 3 or 24 = 4 6, it is usual practice to express a number as a unique product of its prime factors. Using the fact that 24 = 4 6, we can separate 4 and 6 into products of prime factors, namely 2 2 and 2 3. Therefore 24 = (such representations are generally sorted in ascending order, as shown above). In index (power) notation, this is A tree diagram is often helpful when separating numbers into prime factors start by splitting the number into one pair of factors, and keep splitting each factor until you are left only with prime numbers at the end of each branch. Example (6). Separate 60 into its prime factors. We start by seeing that 60 is equal to Neither 6 nor 10 is prime, so we must move down the tree, separating 6 into 2 3 and 10 into 2. Now, all of the resulting factors are prime, so there is no need to go further. We thus have the prime factors of 60 shown in bold at the leaves of the tree. Sorted in ascending order they are In index (power) notation, this is

7 Mathematics Revision Guides Factors, Prime Numbers, H.C.F. and L.C.M. Page 7 of 10 The Highest Common Factor, or H.C.F. The highest common factor of two or more numbers is the largest number that can divide into all of them without a remainder. Example (7) : Andy has bought two large packs of sweets to give to neighbouring children for Trick or Treat. One pack contains 32 lollipops, and the other contains 48 chocolate bars. He wants to ensure that each child has exactly the same number of lollipops and chocolate bars, without any of either left over. What is the largest number of children that could be provided for in this way, and how many lollipops and chocolate bars would each child receive? Andy thinks: Well, there are eight children on our road, and 8 goes into both 32 and 48. That way, each child can have four lollipops and six chocolate bars. In other words, 8 is a common factor of both 32 and 48. He then realises that four lollipops and six chocolate bars can still be split equally between two people since 4 and 6 are both even, and thus he works out that the 32 lollipops and 48 chocolate bars can be split equally between 16 people and not just 8, with one person receiving two lollipops and three chocolate bars. In Example (7), we reasoned that the numbers 32 and 48 had a highest common factor (H.C.F.) of 16. Example (8). Find the H.C.F. of 24 and 54. First, we list the factors of each: Factors of 24: 1, 2, 3, 4, 6, 8, 12, 24 Factors of 54: 1, 2, 3, 6, 9, 18, 27, 54 The H.C.F. of 24 and 54 is the largest factor common to both, namely 6. Example (9). Find the H.C.F. of 90 and 100. First, we list the factors of each: Factors of 90: 1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45, 90 Factors of 100: 1, 2, 4, 5, 10, 20, 25, 50, 100 The H.C.F. of 90 and 100 is 10.

8 Mathematics Revision Guides Factors, Prime Numbers, H.C.F. and L.C.M. Page 8 of 10 Systematic method of finding the H.C.F. If the H.C.F. cannot be found easily, then separate each number into prime factors and find those that are common to both. The lower power of each prime factor becomes a factor of the H.C.F. Any prime factors occurring in only one of the numbers are absent from the H.C.F. Example (9a). Find the H.C.F. of 90 and 100. This H.C.F. can be found to be 10 by inspection, but the formal method is shown below. Separating into prime factors we have Therefore 90 = or , and 100 = or Going through each prime factor in turn: 2 occurs once (as 2 1 ) in 90, but as 2 2 or 2 2 in 100, and so must therefore occur as the lower power, namely 2 1 or simply 2, in the H.C.F. 3 occurs as 3 3 or 3 2 in 90, but not at all in 100, so it cannot be present in the H.C.F. 5 occurs once (as 5 1 ) in 90, but as 5 5 or 5 2 in 100, and so it must occur as 5 1, i.e. 5, in the H.C.F. This result can also be shown by expressing 90 and 100 as prime factor products in long non-index form and highlighting the factors common to both: The H.C.F. of 90 and 100 is therefore 2 5 or 10. This result can also be shown using a Venn diagram with two overlapping regions. All the prime numbers inside the Factors of 90 region multiply together to give 90, and all those inside the Factors of 100 region multiply together to give 100. The factors inside the area of overlap give the H.C.F. when multiplied.

9 Mathematics Revision Guides Factors, Prime Numbers, H.C.F. and L.C.M. Page 9 of 10 The Lowest Common Multiple, or L.C.M. The lowest common multiple of two or more numbers is the smallest number into which all of them can be divided without a remainder. Example (10): Karen is planning a barbecue and wants to buy equal numbers of burgers and buns. Burgers are sold in packs of 12; buns, in packs of 10. What is the smallest total number of burgers and buns she can buy without having any of either left over, and how many packs of each is that? Karen thinks: Well, = 120, and so is This means that if I buy 10 packs of burgers and 12 packs of buns, then I ll have enough to make 120. In other words, 120 is a common multiple of both 10 and 12. She then realises that she might not need 120 burgers and buns after all, since she reckons that 60 could also be divided by 10 and 12. In other words, she buys 5 packs of 12 burgers and 6 packs of 10 buns, making 60 burgers and 60 buns in all. In Example (10), we worked out that 10 and 12 had a lowest common multiple (L.C.M.) of 60. Example (11). Find the L.C.M. 8 and 12. We list the first few multiples of each: Multiples of 8: 8, 16, 24, 32,... Multiples of 12: 12, 24, The L.C.M. of 8 and 12 is the smallest number common to both lists, namely 24. Example (12). Find the L.C.M. of 15 and 18. We list the first few multiples of each: Multiples of 15: 15, 30, 45, 60, 75, 90, 105,... Multiples of 18: 18, 36, 54, 72, 90, 108,... The L.C.M. of 15 and 18 is 90.

10 Mathematics Revision Guides Factors, Prime Numbers, H.C.F. and L.C.M. Page 10 of 10 Systematic method of finding the L.C.M. If the L.C.M. cannot be found easily, then separate each number into prime factors. By contrast with the H.C.F, the L.C.M. must have the highest power of each prime factor to be found in each number. If a factor is present in one number but absent in the other, then it must be included in the L.C.M. Example (13). Find the L.C.M. of 36 and = or , and 60 = or Both 36 and 60 have 2 2 as a factor, therefore it must be in the L.C.M. Again, both numbers have 3 as a factor, but 36 has the higher power (3 2 ), so that must be in the L.C.M. The factor of 5 is absent in 36, but since it is included in 60, it has to be in the L.C.M. as well. The L.C.M. of 36 and 60 is therefore , or 180. This result can also be shown by expressing 36 and 60 as prime factor products in long non-index form and firstly highlighting the factors common to both (thus finding the H.C.F. into the bargain): The H.C.F. of 36 and 60 is 36 = or 12, and we will use this result later. We are still left with unmatched factors of 3 and 5 though. We need to multiply the intermediate H.C.F. of 12 by these unmatched factors of 3 and 5. The L.C.M. of 36 and 60 is therefore , or 180. Venn diagram illustration : We multiply together every factor inside the Venn circles, thus = 180.

### FACTORS, PRIME NUMBERS, H.C.F. AND L.C.M.

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