GAJERA INTERNATIONAL SCHOOL, SURAT SUBJECT: MATHS STUDY MATERIAL DATE: 21/04/2020 STD: 6 Topic:- L..M

Lowest Common Multiple The Lowest Common Multiple (LCM) of two or more given is the lowest (or smallest or least) of their common multiples. METHOD

1. Common multiple method Example:-Find the lowest common multiple of the numbers 8, 12 and 18. Solution : List the multiples of 8: = 8, 16, 24, 32, 40, 48, 56, 64, 72 , 80, 88 List the multiples of 12: = 12, 24, 36, 48, 60, 72 , 84, List the multiples of 18: = 18, 36, 54, 72 , 90, 1 The common multiples of 8, 12 and 18 are 72 L.C.M. of 8, 12 and 18 is 72. This method works only when there are very small numbers.

2. Prime factorization method Example:- Find the LCM of 90, 100 and 150. Solution: Prime factorisation of the given numbers 90 = 2 x 3 x 3 x 5 = 2 x 3 2 x 5 100 = 2 x 2 x 5 x 5 = 2 2 x 5 2 150 = 2 x 3 x 5 x 5 = 2 x 3 x 5 2 Let’s find the of all the factors with highest powers. 22 x 3 2 x 5 2 = 4 x 9 x 25 = 900 or L.C.M = 2× 3 ×2 × 3× 5× 5 = 900 Note:- ( (i) As above example:- 2 is common in all the three factors so we have to take only one time 2 (ii) If 2 is common only in two factor then also we have to 2 only one time)

3. Common division method of prime factorisation A very convenient method to find the LCM is the common division method. In this method of prime factorisation we proceed as follows: Arrange all the given numbers in a row and separated by commas. Start with the lowest prime which divides at least one of the given numbers exactly. Write down the and any undivided numbers in the next line. Repeat the process as shown below until 1 is the only common factor. Find the product of all the . This is the required LCM.

Example1:-Find the L.C.M. of the numbers 36, 48 and 72.

Solution:

LCM = 2 x 2 x 2 x 2 x 3 x 3 = 144

Word problem Find the least number which when divided by 18, 28, 32 and 42 leaves a 5 in each case.

Solution:

First we should find the LCM of 18, 28, 32 and 42.

LCM = 2 x 2 x 2 x 2 x 2 x 3 x 3 x 7 = 2016

2016 is the least number which when divided by the given numbers will leave remainder 0 in each case. But we need the least number that leaves remainder 5 in each case.

Therefore, the required number is 5 more than 2016. The required least number = 2016 + 5 = 2021

Example2 :-The traffic lights at three different road crossings change after every 48 seconds, 72 seconds and 108 seconds respectively. If they change simultaneously at 7 a.m., at what time will they change simultaneously again? Solution: The time period after which these lights will change = LCM of 48, 72 and 108

LCM = 2 x 2 x 2 x 2 x 3 x 3 x 3 = 432 Therefore, the light will change together after every 432 seconds. i.e 7 min 12 seconds. Hence, they will change simultaneously at 7: 07: 12 am.

Some Problems on HCF and LCM We come across a number of situations in which we make use of the concepts of HCF and LCM. We explain these situations through a few examples.

Example 1 : Two tankers contain 850 litres and 680 litres of kerosene oil respectively. Find the maximum capacity of a container which can measure the kerosene oil of both the tankers when used an exact number of times.

Solution : The required container has to measure both the tankers in a way that the count is an exact number of times. So its capacity must be an exact of the capacities of both the tankers. Moreover, this capacity should be maximum . Thus, the maximum capacity of such a container will be the HCF of 850 and 680. It is found as follows : 850 = 2 × 5 × 5 × 17 = 2 × 5 × 17 × 5 680 = 2 × 2 × 2 × 5 × 17 = 2 × 5 × 17 × 2 × 2 The common factors of 850 and 680 are 2, 5 and 17. Thus, the HCF of 850 and 680 is 2 × 5 × 17 = 170. Therefore, maximum capacity of the required container is 170 litres. It will fill the first container in 5 and the second in 4 refills.

Example:-2 In a morning walk, three persons step off together. Their steps measure 80 cm, 85 cm and 90 cm respectively. What is the minimum distance each should walk so that all can cover the same distance in complete steps?

Solution : The distance covered by each one of them is required to be the same as well as minimum. The required minimum distance each should walk would be the lowest common multiple of the measures of their steps. Can you describe why? Thus, we find the LCM of 80, 85 and 90. The LCM of 80, 85 and 90 is 12240. The required minimum distance is 12240 cm.

Example 3 : Find the least number which when divided by 12, 16, 24 and 36 leaves a remainder 7 in each case.

Solution : We first find the LCM of 12, 16, 24 and 36 as follows : 2 12 16 24 36 2 6 8 12 18 2 3 4 6 9 2 3 2 3 9 3 3 1 3 9 3 1 1 1 3 1 1 1 1 Thus, LCM = 2 × 2 × 2 × 2 × 3 × 3 = 144 144 is the least number which when divided by the given numbers will leave remainder 0 in each case. But we need the least number that leaves remainder 7 in each case. Therefore, the required number is 7 more than 144. The required least number = 144 + 7 = 151.

Properties of HCF and LCM Example: 4- Determine the smallest 3-digit number which is exactly divisible by 6, 8 and 12. Solution. Smallest number which is exactly divisible by 6, 8 and 12 = LCM of 6, 8 and 12

∴ LCM of 6, 8 and 12 = 2 x 2 x 2 x 3 = 24

Since, all the multiples of 24 will also be divisible by 6, 8 and 12.

But we need to find the smallest three digit number which is divisible by 6, 8 and 12.

Here 24 × 4 = 96 and 24 × 5 = 120 (after 3 any number multiple With l.C.M of 24 which is divisible by 6,8 and 12)

Thus, 120 is the required smallest three digit number which is exactly divisible by 6, 8 and 12.

HCF and LCM Formulas Property 1: The product of LCM and HCF of any two given natural numbers is equivalent to the product of the given numbers. LCM × HCF = Product of the Numbers Suppose A and B are two numbers, then. LCM (A & B) × HCF (A & B) = A × B Property 2: HCF of co-prime numbers is 1. Therefore LCM of given co -prime numbers is equal to the product of the numbers. LCM of Co-prime Numbers = Prod uct Of The Numbers Property 3: H.C.F. and L.C.M. of LCM of fractions = LCM of numerators / HCF of denominators HCF of fractions = HCF of numerators LCM of denominators

EXAMPLE:- Highest common factor and lowest common multiple of two numbers are 18 and 1782 respectively. One number is 162, find the other number.

Solution:- Given, H.C.F = 18 L.C.M = 1782 First number = 162 Second number = ? Formula = H.C.F. × L.C.M. = First num ber × Second number 18 × 1782 = 162 × Second number 18 × = Second number Therefore, the second number = 198

WORKSHEET

Q- 1 Determine the greatest 3-digit number exactly divisible by 8, 10 and 12. (same as example 4)

Q- 2 Find the smallest 4-digit number which is divisible by 18, 24 and 32. (same as example 4)

Q- 3 Find the LCM of the following numbers: :( using prime factorization method)

(a) 9 and 4 (b) 12 and 5 (c) 6 and 5 (d) 15 and 4

Q-4 Find the lowest common multiple of the numbers:- (a) 12 and 24. (b) 12 and 16 (c) 10 and 15

Q-5 (a) Find (L.C.M) of 50 and 75 by division method.

(b) Find least common multiple (L.C.M) of 120, 144, 160 and 180 by using division method.

(c) Find least common multiple (L.C.M) of 64, and 15 by using division method.

(d) Find least common multiple (L.C.M) of 45, 78, 160 and 180 by using division method

Q-6 Highest common factor and lowest common multiple of two numbers are 15 and 90 respectively. One

number is 30, find the other number.

Q-7 Word problems 1. Three tankers contain 403 litres, 434 litres and 465 litres of diesel respectively. Find the maximum capacity of a container that can measure the diesel of the three containers exact number of times. 2. Find the least number which when divided by 6, 15 and 18 leave remainder 5 in each case. 3. Three boys step off together from the same spot. Their steps measure 63 cm, 70 cm and 77 cm respectively. What is the minimum distance each should cover so that all can cover the distance in complete steps?