Class 8: Factors and Multiples (Lecture Notes)

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1 Class 8: Factors and Multiples (Lecture Notes) If a number a divides another number b exactly, then we say that a is a factor of b and b is a multiple of a. Factor: A factor of a number is an exact divisor of that number. Multiple: A number is said to be a multiple of any of its factors. When 20 is divided by 5, the remainder is zero, i.e. 5 divides 20 exactly. So 5 is a factor of 20 and 20 is a multiple of 5. Even Numbers: All the multiples of 2 are called even numbers. 2, 4, 6, 8, 10, 12 are all even numbers. Odd Numbers: Numbers which are not multiple of 2 are called odd numbers. 1, 3, 5, 7, 9, 11, 13, 15 are all odd numbers. Prime Numbers: A number which has exactly two factors, namely, 1 and the number itself, is called a prime number. 2, 3, 5, 7, 11, 13, 17, 19 are all prime numbers. Composite Number: A number which has more than two factors is called a composite number. 4, 6, 8, 9, 10, 12, 15 are all composite numbers. Note: i. 1 is neither prime nor composite. It is the only factor that has one factor, namely itself. ii. 2 is the smallest prime number. iii. 2 is the only even prime number. Twin Primes: Two consecutives odd prime numbers are known as twin primes. Example of Twin Primes: i. 3, 5 ii. 5, 7 iii. 11, 13 Prime Triplets: A set of three consecutive prime numbers, differing by 2, is called a prime triplet. 1

2 The only prime triplet is {3, 5, 7} Perfect Numbers: If the sum of all the factors of a number is twice the number, then the number is called a perfect number. i. 6 is perfect number. Its factors are 1, 2, 3 and 6. We see that = 12 = (2 x 6) ii. Another example is 28. Its factors are 1, 2, 4, 7, 14 and 28. We see that = 56 = (2 x 28) `Co-Primes: Two numbers are said to be co-primes if they have no common factor other than 1. i. 3, 4 ii. 4, 9 Note: i. Two prime numbers are always co-prime ii. Two co-primes are not necessarily prime numbers. Test of Divisibility 1. Test of Divisibility by 3: A number is divisible by 3 if the sum of the digits is divisible by 3. i. Take a number Add the digits = is divisible by 3. Hence is divisible by 3. ii. Take another number Add the digits = is not completely divisible by 3. Hence is not divisible by Test of Divisibility by 4: A number is divisible by 4 is the number formed by the last two digits is divisible by 4. i. Take a number The last two digits form the number 24. Hence the number is divisible by 4. ii. Take a number The last two digits form the number 41. Hence the number is not divisible by 4. 2

3 3. Test of Divisibility by 5: A number is divisible by 5 if the unit digit is either 0 or 5. i. Take a number The last digit is 5. Hence the number is divisible by 5. ii. Take a number The last digit is 1. Hence the number is not divisible by Test of Divisibility by 6: A number is divisible by 6 if it is divisible both by 2 and 3. i. Take a number The last digit is 2, hence divisible by 2. The sum of the digits is is divisible by 3. Hence the number is divisible both by 2 and 3. Hence the number is divisible by Test of Divisibility by 8: A number is divisible by 8, if the number formed by the last 3 digits is divisible by 8. i. Take a number The number formed by the last three digits is is divisible by 8. Hence the number is divisible by 8. ii. Take a number The number formed by the last three digits is is not divisible by 8. Hence the number is divisible by Test of Divisibility by 9: A number is divisible by 9 if the sum of the digits is divisible by 9. i. Take a number Add the digits = is divisible by 9. Hence is divisible by 9. ii. Take another number Add the digits = is not completely divisible by 9. Hence is not divisible by Test of Divisibility by 10: A number is divisible by 10, if its unit digit is 0. 10, 20, 3430, are all divisible by Test of Divisibility by 11: A number is divisible by 11, if the difference between the sum of its digits at odd places and the sum of the digits at even places is either 0 or a number divisible by 11. 3

4 i. Take a number The sum of the digits at odd places = ( ) = 30. The sum of the digits at even places = ( ) = 8. The difference is 22 which is divisible by 11. Hence the number is divisible by 11. ii. Take a number The sum of the digits at odd places = ( ) = 29. The sum of the digits at even places = ( ) = 8. The difference is 21 which is not divisible by 11. Hence the number is not divisible by 11. Prime Factors: A factor of a given number is called a prime factor if this is a prime factor. All the factors of 21 are 1, 3, 7 and 21. Of these, 3 and 7 are prime numbers. Therefore 3 and 7 are prime factors of 21. Prime Factorization: The method of expressing a natural number as a product of prime numbers is call prime factorizations or complete factorization of a given number. Unique Factorizations Property: Any natural number can be expressed as a product of a unique collection of prime numbers except for the order of these prime numbers. The prime factorization of 36 by division method Therefore 36 = 2 x 2 x 3 x 3 = 2 2 x 3 2 Common Factors: A number which divides each one of the given numbers exactly, is called a common factor of each of the given numbers. The factors of 36 are 1, 2, 3, 4, 6, 9, 18 and 36. The factors of 42 are 1, 2, 3, 6, 7, 14, 21 and 42. Hence the common factors of 36 and 42 are 1, 2, 3 and 6. Highest Common Factor (H.C.F) or Greatest Common Divisor (G.C.D): The H.C.F or G.D.C of two or more numbers is the greatest number that divides each of the given numbers exactly. 4

5 Let s take two numbers 36 and 54. The factors of 36 are 1, 2, 3, 4, 6, 9, 18 and 36. The factors of 54 are 1, 2, 3, 6, 9, 18, 27 and 54. Therefore the HCF or 36 and 54 is 18. Method of finding the H.C.F of a given numbers 1. Prime Factorization Method i. Step 1: Express each one of the given numbers as the product of prime factors ii. Step 2: The product of terms containing least powers of common prime factors gives the H.C.F. of the given numbers. H.C.F of 324, 288 and 360. First represent each of the given numbers in to prime factors 324 = 2 2 x = 2 5 x = 2 3 x 3 2 x 5 Therefore H.C.F = Product of terms containing the least powers of common prime factors = 2 2 x 3 2 = Long Division Method i. Divide the larger number by smaller number ii. Divide the divisor by smaller one iii. Repeat the process of dividing the preceding divisor by the remainder last obtained, till remainder 0 is obtained. Example 1: Find H.C.F of 1824 and Example 2: Find H.C.F of 1302, 3465 and 7350 First find the H.C.F 7350 and

6 The find the H.C.F.105 and Therefore H.C.F of 1302, 3465 and 7350 is 21. Applications of H.C.F 1. Co-Prime Numbers: Two natural numbers are said to be co-prime, if their H.C.F is To reduce a fraction (p/q) to the simplest form (lowest form), divide each of the numerator and denominator by the H.C.F of p and q. Least Common Multiple (L.C.M): The L.C.M of two or more numbers is the least natural number which is a multiple of each of the given numbers. Take two numbers say 12 and 18 Multiple of 12 = 12, 24, 36, 48 Multiple of 18 = 18, 36, 54 Hence the L.C.M of 12 and 18 is 36 Methods of finding L.C.M of given numbers 1. Prime Factorization Method i. Express each one of the numbers as the product of prime factors 6

7 ii. The product of all the different prime factors each raised to highest power that appears in the prime factorization of any given numbers, gives the L.C.M of the given numbers. Find L.C.M of 180, 216 and 324 Convert each of the numbers into prime factors. 180 = 2 2 x 3 2 x = 2 2 x = 2 2 x 3 4 Therefore L.C.M = 2 3 x 3 4 x 5 = Common Division Method i. Arrange the numbers in any order ii. Divide by a number that divides exactly at least two of the given numbers and carry forward the other numbers iii. Repeat Step 2 till no two numbers are divisible by the same number, other than 1 iv. The product of the divisors and the un-divided numbers is the required L.C.M Find L.C.M of 16, 18 and Hence the L.C.M = 2 x 2 x 2 x 3 x 2 x 3 = 144 Relation between H.C.F and L.C.M of two numbers i. Product of two given numbers = Product of their H.C.F and L.C.M ii. H.C.F and L.C.M of fractions a. H.C.F of a given fractions = (H.C.F of numerators / L.C.M of the denominators) b. L.C.M of a given fractions = (L.C.M of numerators / H.C.F of the denominators) 7

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