Std 6 Mathematics · Chapter 5

5. GCD and LCM

Solutions to Practice Set 5.1, 5.2, 5.3 and Exercise 5. Maharashtra State Board (Balbharati), Grade Six Mathematics.

About the chapter

In this chapter, we learn to write a number as a product of prime factors. We find all the divisors of a number. Then we find the greatest common divisor (GCD) and the least common multiple (LCM) of numbers. We use GCD and LCM to solve word problems.

Key points

Every composite number can be written as a product of prime numbers.
The greatest number that divides the given numbers exactly is their GCD.
The smallest number that is a multiple of all the given numbers is their LCM.
The GCD of two coprime numbers is 1.
If something is divided into the largest equal parts, find the GCD. If events repeat and meet again, find the LCM.

Formulas and terms

Prime factorisation – writing a number as a product of prime numbers. 12 = 2 × 2 × 3
GCD – Greatest Common Divisor. It is also called HCF.
LCM – Least Common Multiple.
Product of two numbers = GCD × LCM

Practice Set 5.1

1Write each number as a product of prime factors.

(i) 69 = 3 × 23
(ii) 27 = 3 × 9 = 3 × 3 × 3
(iii) 78 = 2 × 39 = 2 × 3 × 13
(iv) 126 = 2 × 63 = 2 × 7 × 9 = 2 × 3 × 3 × 7

Factor tree for 126 — keep splitting until every branch ends in a prime number.

2Write all the divisors of each of the following numbers.

(i) 18 – 1, 2, 3, 6, 9, 18
(ii) 56 – 1, 2, 4, 7, 8, 14, 28, 56
(iii) 120 – 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60, 120
(iv) 256 – 1, 2, 4, 8, 16, 32, 64, 128, 256
(v) 67 – 1, 67 (67 is a prime number.)

Practice Set 5.2

1Find the GCD of the following numbers.

(i) 36, 48

36 = 2 × 2 × 3 × 3
48 = 2 × 2 × 2 × 2 × 3
GCD = 2 × 2 × 3 = 12

(ii) 31, 2

31 and 2 are both prime numbers. Their only common divisor is 1.
GCD = 1

(iii) 65, 13

65 = 5 × 13
13 = 1 × 13
GCD = 13

(iv) 108, 132

108 = 2 × 2 × 3 × 3 × 3
132 = 2 × 2 × 3 × 11
GCD = 2 × 2 × 3 = 12

(v) 30, 24, 48

30 = 2 × 3 × 5
24 = 2 × 2 × 2 × 3
48 = 2 × 2 × 2 × 2 × 3
GCD = 2 × 3 = 6

Divisor method for (i): the greatest number in the common part is the GCD.

2The length and breadth of a room are 15 m and 9 m. What is the maximum length of each square tile to be laid in this room?

The maximum length of the tile is the GCD of 15 and 9.

Divisors of 15 : 1, 3, 5, 15
Divisors of 9 : 1, 3, 9
GCD = 3

The maximum length of each square tile is 3 metres.

3A shopkeeper has two rolls of cloth of length 90 m and 120 m. He wants to cut both rolls into pieces of equal maximum length. What can be the maximum length of each piece?

The maximum length of each piece is the GCD of 90 and 120.

90 = 2 × 3 × 3 × 5
120 = 2 × 2 × 2 × 3 × 5
GCD = 2 × 3 × 5 = 30

The maximum length of each piece is 30 metres.

4A garland seller has 48 red and 60 white roses. He wants to make separate garlands with the same number of flowers in each. What is the maximum number of flowers in each garland?

The maximum number of flowers is the GCD of 48 and 60.

48 = 2 × 2 × 2 × 2 × 3
60 = 2 × 2 × 3 × 5
GCD = 2 × 2 × 3 = 12

The maximum number of flowers in each garland is 12.

Practice Set 5.3

1Find the LCM of the following numbers.

(i) 16, 24

16 = 2 × 2 × 2 × 2
24 = 2 × 2 × 2 × 3
LCM = 2 × 2 × 2 × 2 × 3 = 48

(ii) 30, 40

30 = 2 × 3 × 5
40 = 2 × 2 × 2 × 5
LCM = 2 × 2 × 2 × 3 × 5 = 120

(iii) 72, 96

72 = 2 × 2 × 2 × 3 × 3
96 = 2 × 2 × 2 × 2 × 2 × 3
LCM = 2 × 2 × 2 × 2 × 2 × 3 × 3 = 288

(iv) 9, 15, 20

9 = 3 × 3
15 = 3 × 5
20 = 2 × 2 × 5
LCM = 2 × 2 × 3 × 3 × 5 = 180

(v) 30, 45, 60

30 = 2 × 3 × 5
45 = 3 × 3 × 5
60 = 2 × 2 × 3 × 5
LCM = 2 × 2 × 3 × 3 × 5 = 180

2Solve the following examples.

(i) Find the smallest three-digit number divisible by 5, 10 and 15.

LCM of 5, 10 and 15 = 30
Multiples of 30 : 30, 60, 90, 120, …
The smallest three-digit multiple is 120.

The smallest three-digit number is 120.

(ii) The product of two numbers is 360 and their GCD is 6. What is their LCM?

Product of two numbers = GCD × LCM
360 = 6 × LCM
LCM = 360 ÷ 6 = 60

The LCM is 60.

(iii) The LCM of two numbers is 200 and the GCD is 10. If one number is 40, what is the other?

First number × Second number = GCD × LCM
40 × Second number = 10 × 200 = 2000
Second number = 2000 ÷ 40 = 50

The other number is 50.

(iv) A farmer has 250 kg of wheat and 150 kg of jowar to be filled in sacks of the same weight. What is the maximum weight in each sack?

250 = 2 × 5 × 5 × 5
150 = 2 × 3 × 5 × 5
GCD = 2 × 5 × 5 = 50

The maximum weight in each sack is 50 kg.

(v) Three friends take 2, 3 and 4 minutes for one lap. After how many minutes will they meet again at the starting place?

Multiples of 2 : 2, 4, 6, 8, 10, 12, …
Multiples of 3 : 3, 6, 9, 12, …
Multiples of 4 : 4, 8, 12, …
LCM = 12

They will meet again after 12 minutes.

(vi) Children stand 6 in each row or 8 in each row with no child left out. What is the minimum number of children?

Multiples of 6 : 6, 12, 18, 24, …
Multiples of 8 : 8, 16, 24, …
LCM = 24

The minimum number of children is 24.

(vii) Three buses leave every 10, 15 and 20 minutes, first at 8 a.m. When will they leave together again?

10 = 2 × 5
15 = 3 × 5
20 = 2 × 2 × 5
LCM = 2 × 2 × 3 × 5 = 60 minutes = 1 hour

The buses will leave together again at 9 a.m.

Exercise 5

1Choose the correct option.

(i) What is the smallest prime factor of 75? — b) 3
75 = 3 × 5 × 5
(ii) What is the least common divisor of 17 and 18? — a) 1
1 is a divisor of every number.

2State True or False.

(i) Every prime number has only two factors. – True
(ii) The product of two prime numbers is always a prime number. – False
2 × 3 = 6, and 6 is not a prime number.
(iii) GCD is a divisor of LCM. – True
(iv) LCM is greater than the given numbers, or equal to the smallest of them. – False
The LCM is equal to or greater than the largest of the given numbers.

Note on (iv): If your textbook says "equal to the largest of them", then the answer is True.

3Express the number as a product of prime factors.

(i) 91 = 7 × 13
(ii) 160 = 2 × 80 = 2 × 2 × 40 = 2 × 2 × 2 × 20 = 2 × 2 × 2 × 2 × 10 = 2 × 2 × 2 × 2 × 2 × 5
(iii) 89 = 1 × 89 (89 is a prime number.)

4Find the GCD of the following numbers.

(i) 21, 42

21 = 3 × 7
42 = 2 × 3 × 7
GCD = 3 × 7 = 21

(ii) 17, 9

17 = 1 × 17
9 = 3 × 3
The only common divisor is 1. GCD = 1

(iii) 120, 180, 240

120 = 2 × 2 × 2 × 3 × 5
180 = 2 × 2 × 3 × 3 × 5
240 = 2 × 2 × 2 × 2 × 3 × 5
GCD = 2 × 2 × 3 × 5 = 60

5Find the LCM of the following numbers.

(i) 11, 16

11 = 11
16 = 2 × 2 × 2 × 2
LCM = 2 × 2 × 2 × 2 × 11 = 176

(ii) 7, 49

7 = 7
49 = 7 × 7
LCM = 7 × 7 = 49

(iii) 16, 48, 6

16 = 2 × 2 × 2 × 2
48 = 2 × 2 × 2 × 2 × 3
6 = 2 × 3
LCM = 2 × 2 × 2 × 2 × 3 = 48

6Solve.

(i) The product of two numbers is 120. If their LCM is 60, what is their GCD?

Product of two numbers = GCD × LCM
120 = GCD × 60
GCD = 120 ÷ 60 = 2

The GCD is 2.

(ii) The GCD of a number and 16 is 4 and their LCM is 48. What is that number?

Number × 16 = GCD × LCM
Number × 16 = 4 × 48 = 192
Number = 192 ÷ 16 = 12

The number is 12.

(iii) A carpenter cuts two strips of 54 inches and 72 inches into equal lengths with nothing left over. What is the maximum length of a strip?

54 = 2 × 3 × 3 × 3
72 = 2 × 2 × 2 × 3 × 3
GCD = 2 × 3 × 3 = 18

The maximum length of each strip is 18 inches.

(iv) Samip has 84 mangoes and 105 apples to be packed in boxes with the same number of fruits, each box holding only one kind of fruit. (1) What is the maximum number of fruits in a box? (2) How many boxes are needed?

84 = 2 × 2 × 3 × 7
105 = 3 × 5 × 7
GCD = 3 × 7 = 21

(1) The maximum number of fruits in a box is 21.

Boxes of mangoes = 84 ÷ 21 = 4
Boxes of apples = 105 ÷ 21 = 5
Total boxes = 4 + 5 = 9

(2) He needs 9 boxes.

✦Mathematical puzzle – solve the two grids like the given example.

Fill the empty cells with prime numbers. The product of the numbers in each row and each column must be the number given at its end.

Puzzle (i): rows 8, 18, 30 and columns 8, 18, 30

Puzzle (ii): rows 42, 45, 66 and columns 12, 27, 385

Puzzle (i)
Rows : 2 × 2 × 2 = 8, 2 × 3 × 3 = 18, 2 × 3 × 5 = 30
Columns : 2 × 2 × 2 = 8, 2 × 3 × 3 = 18, 2 × 3 × 5 = 30

Puzzle (ii)
Rows : 2 × 3 × 7 = 42, 3 × 3 × 5 = 45, 2 × 3 × 11 = 66
Columns : 2 × 3 × 2 = 12, 3 × 3 × 3 = 27, 7 × 5 × 11 = 385

✦Think about.

(1) If the GCD increases, does the LCM increase?

No, not always. For two numbers, GCD × LCM = product of the numbers. For 4 and 6, GCD = 2 and LCM = 12. For 4 and 8, GCD = 4 and LCM = 8. Here the GCD increased, but the LCM decreased.

(2) How do you decide whether to calculate the GCD or the LCM in a word problem?

When a thing is divided into the largest equal parts, we find the GCD. When events repeat and happen together again, we find the LCM.

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