Finding Common Ground Class 7 Free Notes and Mind Map (Free PDF Download)

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This chapter explains how to find the Highest Common Factor (HCF) and Lowest Common Multiple (LCM) of numbers using prime factorisation. You will learn efficient methods, properties, patterns, and real-life applications of HCF and LCM.

The Greatest of All

Sameeksha’s Tile Problem

  • Sameeksha is building her new house with a main room measuring 12 ft by 16 ft.
  • She wants to cover the floor with square tiles of the same size, using as few tiles as possible, with tile length being a whole number of feet.
  • For tiles to fit the breadth exactly, the side of the tile should be a factor of 12.
  • For tiles to fit the length exactly, the side should be a factor of 16.
  • So the side of the tile should be a common factor of both 12 and 16.
  • Factors of 12 are: 1, 2, 3, 4, 6, 12.
  • Factors of 16 are: 1, 2, 4, 8, 16.
  • Common factors are: 1, 2, and 4.
  • Square tiles can have sides 1 ft, 2 ft, and 4 ft, but she should use the largest sized square tile (4 ft) to minimize the number of tiles.
  • Therefore, she needs tiles of size 4 ft.

Questions:

How many tiles of this size should she purchase?

  • Solution: Room area = 12 × 16 = 192 sq ft. Each tile area = 4 × 4 = 16 sq ft. Number of tiles = 192 ÷ 16 = 12 tiles.

What if Sameeksha did not insist on the length of the tile to be a whole number of feet and the length could be a fractional number of feet? Would the answer change?

  • Solution: If fractional lengths are allowed, she could use even larger tiles (like 12 ft or 16 ft), but practically tiles are whole numbers.

Definition of HCF

  • The Highest Common Factor (HCF) of two or more numbers is the highest (or greatest) of their common factors.
  • It is also known as the Greatest Common Divisor (GCD).
  • In the previous problem, 4 is the HCF of 12 and 16.

Lekhana’s Rice Packing Problem

  • Lekhana purchases rice from two farms: 84 kg from one farm and 108 kg from the other.
  • She wants rice to be packed in bags, each bag has rice from only one farm, all bags have the same weight that is a whole number of kg.
  • If she wants to use as few bags as possible, what should the weight of each bag be?
  • To divide 84 kg into bags of equal weight, we need the factors of 84.
  • Similarly, for 108, we need the factors of 108.
  • Factors of 84: 1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42, and 84.
  • Factors of 108: 1, 2, 3, 4, 6, 9, 12, 18, 27, 36, 54, and 108.
  • Common factors of 84 and 108 are: 1, 2, 3, 4, 6, and 12.
  • To minimise the number of bags, she should choose the largest common factor, which is 12 kg.

Jump Jackpot Game

Question: In the Jump Jackpot game, treasures are placed on two numbers. Find the longest jump size (starting from 0) using which Jumpy can land on both numbers.

(a) 14 and 30

  • Solution: Factors of 14 = 1, 2, 7, 14. Factors of 30 = 1, 2, 3, 5, 6, 10, 15, 30. Common factors = 1, 2. Longest jump size = 2 (which is HCF).

(b) 7 and 11

  • Solution: Factors of 7 = 1, 7. Factors of 11 = 1, 11. Common factors = 1. Longest jump size = 1 (which is HCF).

(c) 30 and 50

  • Solution: Factors of 30 = 1, 2, 3, 5, 6, 10, 15, 30. Factors of 50 = 1, 2, 5, 10, 25, 50. Common factors = 1, 2, 5, 10. Longest jump size = 10 (which is HCF).

(d) 28 and 42

  • Solution: Factors of 28 = 1, 2, 4, 7, 14, 28. Factors of 42 = 1, 2, 3, 6, 7, 14, 21, 42. Common factors = 1, 2, 7, 14. Longest jump size = 14 (which is HCF).

Is the longest jump size for the numbers the same as their HCF? Explain why it is so.

  • Solution: Yes, the longest jump size is always the HCF. This is because if you can jump by the HCF, you will land on all multiples of the HCF, which include both numbers (since both are multiples of their HCF).

Primes

  • A prime is a number greater than 1 that has only 1 and the number itself as its factors.
  • Examples: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, etc.

Prime Factorisation

  • Any number can be written as a product of primes — keep rewriting composite factors until only primes are left.
  • For example, prime factorisation of 90: 90 = 3 × 3 × 2 × 5 = 2 × 3² × 5.
  • The number 90 could also have been factorised as 3 × 30 or 2 × 45, but all lead to the same prime factors with perhaps a change in their order.
  • Prime factorisation of a prime number is the prime number itself.

Procedure for Prime Factorisation (Division Method)

  • Divide the number by the smallest prime factor repeatedly until you reach 1.
  • Example: Prime factorisation of 105:
text3 | 105
5 | 35
7 | 7
    1

105 = 3 × 5 × 7
  • Example: Prime factorisation of 30:
text2 | 30
3 | 15
5 | 5
    1

30 = 2 × 3 × 5

Question: Try finding the prime factorisation of 1200 using the division method.

Solution:

2 | 1200
2 | 600
2 | 300
2 | 150
3 | 75
5 | 25
5 | 5
    1

1200 = 2 × 2 × 2 × 2 × 3 × 5 × 5 = 2⁴ × 3 × 5²

Factors of a Number Using Prime Factorisation

  • From the prime factorisation of a number, we can construct all its factors.
  • Consider the number 840 and its prime factorisation: 840 = 2 × 2 × 2 × 3 × 5 × 7.

Is 2 × 2 × 7 = 28 a factor of 840?

  • Solution: Yes, 28 is a factor. Reordering: 840 = (2 × 2 × 7) × (2 × 3 × 5) = 28 × 30.

If yes, what should it be multiplied by to get 840?

  • Solution: It should be multiplied by 30 (which is 2 × 3 × 5).

Is 2 × 7 = 14 a factor of 840? Why or why not?

  • Solution: Yes, because 14 can be formed from a subpart of the prime factorisation (2 × 7).

Is 2 × 2 × 2 a factor of 840? Why or why not?

  • Solution: Yes, because 2 × 2 × 2 = 8 can be formed from the prime factorisation.

Is 3 × 3 × 3 a factor of 840? Why or why not?

  • Solution: No, because there is only one 3 in the prime factorisation of 840, not three 3s.

Finding All Factors of 225 Using Prime Factorisation

Question: Find the factors of 225 using prime factorisation.

Solution:

5 | 225
5 | 45
3 | 9
3 | 3
    1

225 = 5 × 5 × 3 × 3 = 5² × 3²
  • Prime factors: 3, 5.
  • Combination of two prime factors: 3 × 3 = 9, 5 × 5 = 25, 3 × 5 = 15.
  • Combination of three prime factors: 3 × 3 × 5 = 45, 3 × 5 × 5 = 75.
  • Combination of four prime factors: 3 × 3 × 5 × 5 = 225.
  • Adding 1 to this list, the factors of 225 are: 1, 3, 5, 9, 15, 25, 45, 75, 225.

Figure it Out

Question: List all the factors of the following numbers:

(a) 90

Solution:

2 | 90
3 | 45
3 | 15
5 | 5
    1

90 = 2 × 3 × 3 × 5 = 2 × 3² × 5
  • Factors of 90: 1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45, 90.

(b) 132

Solution:

2 | 132
2 | 66
3 | 33
11| 11
    1

132 = 2 × 2 × 3 × 11 = 2² × 3 × 11
  • Factors of 132: 1, 2, 3, 4, 6, 11, 12, 22, 33, 44, 66, 132.

(c) 360 (this number has 24 factors)

Solution:

2 | 360
2 | 180
2 | 90
3 | 45
3 | 15
5 | 5
    1

360 = 2 × 2 × 2 × 3 × 3 × 5 = 2³ × 3² × 5
  • Factors of 360: 1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 18, 20, 24, 30, 36, 40, 45, 60, 72, 90, 120, 180, 360 (24 factors).

Anshu’s Claim (Conjecture)

Anshu’s claim: “The larger a number is, the longer its prime factorisation will be.”

What do you think of Anshu’s claim?

  • Solution: Anshu’s claim is not always true. For example:
    • 96 = 2 × 2 × 2 × 2 × 2 × 3 (6 prime factors).
    • 121 = 11 × 11 (2 prime factors).
    • Here, 121 > 96, but 121 has a shorter prime factorisation.
  • This is called a conjecture (a statement made without proof). We disproved this conjecture by finding a counterexample.

Finding the HCF of Numbers Using Prime Factorisation

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

Solution:

  • Prime factorisation:
    • 45 = 3 × 3 × 5
    • 75 = 3 × 5 × 5
  • Common subparts (factors common to both):
    • 3 (appears in both)
    • 5 (appears in both)
    • 3 × 5 = 15 (appears in both)
  • Common factors: 1, 3, 5, 15.
  • HCF = 15 (highest among common factors).

Example 2: Find the common factors and the HCF of 112 and 84

Solution:

  • Prime factorisation:
    • 112 = 2 × 2 × 2 × 2 × 7 = 2⁴ × 7
    • 84 = 2 × 2 × 3 × 7 = 2² × 3 × 7
  • Common prime factors: 2 and 7.
  • Minimum occurrences:
    • 2 appears 4 times in 112 and 2 times in 84, so take 2 times (minimum).
    • 7 appears 1 time in both, so take 1 time.
  • HCF = 2 × 2 × 7 = 28.

Example 3: Find the common factors and the HCF of 30 and 72

Solution:

  • Prime factorisation:
    • 30 = 2 × 3 × 5
    • 72 = 2 × 2 × 2 × 3 × 3 = 2³ × 3²
  • Common prime factors: 2 and 3.
  • Minimum occurrences:
    • 2 appears 1 time in 30 and 3 times in 72, so take 1 time.
    • 3 appears 1 time in 30 and 2 times in 72, so take 1 time.
  • HCF = 2 × 3 = 6.

Example 4: Find the HCF of 225 and 750

Solution:

  • Prime factorisation:
    • 225 = 3 × 3 × 5 × 5 = 3² × 5²
    • 750 = 2 × 3 × 5 × 5 × 5 = 2 × 3 × 5³
  • Common prime factors: 3 and 5.
  • Minimum occurrences:
    • 3 appears 2 times in 225 and 1 time in 750, so take 1 time.
    • 5 appears 2 times in 225 and 3 times in 750, so take 2 times.
  • HCF = 3 × 5 × 5 = 75.

Figure it Out

Question 1: Find the HCF of the following numbers:

(a) 24, 180

Solution:

  • Prime factorisation:
    • 24 = 2 × 2 × 2 × 3 = 2³ × 3
    • 180 = 2 × 2 × 3 × 3 × 5 = 2² × 3² × 5
  • Common primes: 2, 3.
  • Minimum occurrences: 2² × 3 = HCF = 12.

(b) 42, 75, 24

Solution:

  • Prime factorisation:
    • 42 = 2 × 3 × 7
    • 75 = 3 × 5 × 5 = 3 × 5²
    • 24 = 2 × 2 × 2 × 3 = 2³ × 3
  • Common prime: 3 (appears in all three).
  • Minimum occurrences: 3 appears once in each.
  • HCF = 3.

(c) 240, 378

Solution:

  • Prime factorisation:
    • 240 = 2 × 2 × 2 × 2 × 3 × 5 = 2⁴ × 3 × 5
    • 378 = 2 × 3 × 3 × 3 × 7 = 2 × 3³ × 7
  • Common primes: 2, 3.
  • Minimum occurrences: 2¹ × 3¹ = HCF = 6.

(d) 400, 2500

Solution:

  • Prime factorisation:
    • 400 = 2 × 2 × 2 × 2 × 5 × 5 = 2⁴ × 5²
    • 2500 = 2 × 2 × 5 × 5 × 5 × 5 = 2² × 5⁴
  • Common primes: 2, 5.
  • Minimum occurrences: 2² × 5² = 4 × 25 = HCF = 100.

(e) 300, 800

Solution:

  • Prime factorisation:
    • 300 = 2 × 2 × 3 × 5 × 5 = 2² × 3 × 5²
    • 800 = 2 × 2 × 2 × 2 × 2 × 5 × 5 = 2⁵ × 5²
  • Common primes: 2, 5.
  • Minimum occurrences: 2² × 5² = 4 × 25 = HCF = 100.

Question 2: Consider the numbers 72 and 144. Suppose they are factorised into composite numbers as: 72 = 6 × 12 and 144 = 8 × 18. Seeing this, can one say that these two numbers have no common factor other than 1? Why not?

Solution:

  • No, we cannot say that. Even though 6, 12, 8, and 18 may not appear to have obvious common factors, when we do prime factorisation, we find:
    • 72 = 2³ × 3²
    • 144 = 2⁴ × 3²
  • HCF = 2³ × 3² = 8 × 9 = 72.
  • So, 72 and 144 have many common factors, not just 1.

Least, but not Last!

Anshu and Guna’s Toran Problem

  • Anshu and Guna make torans (decorative strips) out of cloth.
  • Anshu uses strips of 6 cm length and Guna uses strips of 8 cm length.
  • If both have to make torans of the same length, what is the smallest possible length?
  • Anshu’s torans can be: 6, 12, 18, 24, 30, 36, 42, 48, 54 cm, … (multiples of 6).
  • Guna’s torans can be: 8, 16, 24, 32, 40, 48, 56, 64, 72 cm, … (multiples of 8).
  • Common multiples of 6 and 8 are: 24, 48, 72, …
  • The smallest common multiple is 24 cm.
  • Therefore, the shortest toran that both can stitch is 24 cm.

Kabamai’s Gajak Problem

  • A sweet shop gives out free gajak to school children on Mondays.
  • Today is a Monday, but Kabamai visits the sweet shop once every 10 days.
  • When is the next time she would be able to get free gajak?
  • Shop gives free sweets every Monday: 7, 14, 21, 28, 35, 42, 49, 56, 63, 70, 77, … days (multiples of 7).
  • Kabamai visits: 10, 20, 30, 40, 50, 60, 70, … days (multiples of 10).
  • Common multiples of 7 and 10: 70, 140, …
  • The first common multiple is 70 days.
  • Therefore, Kabamai will eat free gajak again after 70 days.

Definition of LCM

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

Idli-Vada Game

Question: In the Idli-Vada game, two numbers are chosen. Find the first number for which ‘idli-vada’ will be called out (i.e., find the LCM):

(a) 4 and 6

  • Solution: Multiples of 4: 4, 8, 12, 16, 20, 24, …
  • Multiples of 6: 6, 12, 18, 24, …
  • LCM = 12.

(b) 7 and 11

  • Solution: Multiples of 7: 7, 14, 21, 28, 35, 42, 49, 56, 63, 70, 77, …
  • Multiples of 11: 11, 22, 33, 44, 55, 66, 77, …
  • LCM = 77.

(c) 14 and 30

  • Solution:
    • 14 = 2 × 7
    • 30 = 2 × 3 × 5
    • LCM = 2 × 3 × 5 × 7 = 210.

(d) 15 and 55

  • Solution:
    • 15 = 3 × 5
    • 55 = 5 × 11
    • LCM = 3 × 5 × 11 = 165.

Is the answer always the LCM of the two numbers? Explain.

  • Solution: Yes, because ‘idli-vada’ is called when the number is a multiple of both numbers, and the first such number is the LCM.

Finding LCM through Prime Factorisation

Comparing Prime Factorisations

  • Every factor of a number is formed by taking a subpart of its prime factorisation.
  • Every multiple of a number contains the prime factorisation of that number.
  • Example: 36 = 2 × 2 × 3 × 3, and 648 = 36 × 18 = (2 × 2 × 3 × 3) × (2 × 3 × 3).
  • The prime factors of the multiple contain the prime factors of the number along with some more.

Example 6: Find the LCM of 14 and 35

Solution:

  • Prime factorisation:
    • 14 = 2 × 7
    • 35 = 5 × 7
  • Common primes: 7.
  • All primes: 2, 5, 7.
  • Maximum occurrences:
    • 2 appears 1 time (in 14).
    • 5 appears 1 time (in 35).
    • 7 appears 1 time (in both).
  • LCM = 2 × 5 × 7 = 70.

Is 2 × 3 × 5 × 7 also a common multiple?

  • Solution: Yes, it is a common multiple, but not the lowest common multiple.

Example 7: Find the LCM of 96 and 360

Solution:

  • Prime factorisation:
    • 96 = 2 × 2 × 2 × 2 × 2 × 3 = 2⁵ × 3
    • 360 = 2 × 2 × 2 × 3 × 3 × 5 = 2³ × 3² × 5
  • All primes: 2, 3, 5.
  • Maximum occurrences:
    • 2 appears 5 times (in 96).
    • 3 appears 2 times (in 360).
    • 5 appears 1 time (in 360).
  • LCM = 2⁵ × 3² × 5 = 32 × 9 × 5 = 1440.

Figure it Out

Question: Find the LCM of the following numbers:

(a) 30, 72

Solution:

  • Prime factorisation:
    • 30 = 2 × 3 × 5
    • 72 = 2³ × 3²
  • All primes: 2, 3, 5.
  • Maximum occurrences: 2³ × 3² × 5 = 8 × 9 × 5 = LCM = 360.

(b) 105, 195, 65

Solution:

  • Prime factorisation:
    • 105 = 3 × 5 × 7
    • 195 = 3 × 5 × 13
    • 65 = 5 × 13
  • All primes: 3, 5, 7, 13.
  • Maximum occurrences: 3 × 5 × 7 × 13 = LCM = 1365.

Patterns, Properties, and a Pretty Procedure!

Patterns in HCF

Observation: The HCF of 6 and 18 is 6, which is one of the two given numbers.

Question: Find more such number pairs where the HCF is one of the two numbers. How can we describe such pairs of numbers?

Solution:

  • This happens when one number is a factor of the other.
  • Examples: (5, 10), (7, 21), (12, 36).
  • Generalisation: If n is a number, then any multiple of n can be written as a positive integer multiplied by n. For n and 5n, the HCF = n.

Question: For number pairs satisfying this property:

(a) If m is a number, what could be the other number?

  • Solution: The other number could be any multiple of m (like 2m, 3m, 4m, km, etc.).

(b) If 7k is a number, what could be the other number?

  • Solution: The other number could be any multiple of 7k (like 14k, 21k, 35k, etc.).

Figure it Out

Question 1: Make a general statement about the HCF for the following pairs of numbers:

(a) Two consecutive even numbers

  • Solution: HCF of two consecutive even numbers is 2. Examples: HCF(2, 4) = 2, HCF(8, 10) = 2, HCF(14, 16) = 2.

(b) Two consecutive odd numbers

  • Solution: HCF of two consecutive odd numbers is 1. Examples: HCF(3, 5) = 1, HCF(9, 11) = 1, HCF(15, 17) = 1.

(c) Two even numbers

  • Solution: HCF of two even numbers is at least 2. The exact HCF depends on the numbers.

(d) Two consecutive numbers

  • Solution: HCF of two consecutive numbers is always 1. Examples: HCF(5, 6) = 1, HCF(10, 11) = 1.

(e) Two co-prime numbers

  • Solution: HCF of two co-prime numbers is 1 (by definition, co-prime means no common factors except 1).

Question 2: The LCM of 3 and 24 is 24 (it is one of the two given numbers).

(a) Find more such number pairs where the LCM is one of the two numbers.

  • Solution: Examples: (4, 12), (5, 15), (6, 18), (7, 28).

(b) Make a general statement about such numbers. Describe such number pairs using algebra.

  • Solution: This happens when one number is a multiple of the other. If n and kn are two numbers (where k is a positive integer), then LCM = kn.

Question 3: Make a general statement about the LCM for the following pairs of numbers:

(a) Two multiples of 3

  • Solution: LCM of two multiples of 3 is also a multiple of 3.

(b) Two consecutive even numbers

  • Solution: LCM of two consecutive even numbers is their product divided by 2. Example: LCM(2, 4) = 4, LCM(6, 8) = 24 = (6 × 8) / 2.

(c) Two consecutive numbers

  • Solution: LCM of two consecutive numbers is their product (because HCF = 1). Example: LCM(5, 6) = 30, LCM(7, 8) = 56.

(d) Two co-prime numbers

  • Solution: LCM of two co-prime numbers is their product. Example: LCM(3, 7) = 21, LCM(5, 11) = 55.

Effect of Doubling on HCF

Question: What happens to the HCF of two numbers if both numbers are doubled?

Solution:

  • If both numbers are doubled, the HCF also doubles.
  • Example: Consider 270 and 50.
    • 270 = 2 × 3³ × 5
    • 50 = 2 × 5²
    • HCF = 2 × 5 = 10.
  • Now double them: 540 and 100.
    • 540 = 2² × 3³ × 5
    • 100 = 2² × 5²
    • HCF = 2² × 5 = 20 (which is double of the original HCF).

Explanation: If both numbers are doubled, both get an extra factor of 2 in their prime factors. This 2 will be included in the largest common subpart, so the HCF will double.

Question: Here are some numbers where both are multiples of the same number. Find their HCF:

(a) 18 × 10, 18 × 15

  • Solution: HCF = 18 × HCF(10, 15) = 18 × 5 = 90.

(b) 10 × 38, 10 × 21

  • Solution: HCF = 10 × HCF(38, 21) = 10 × 1 = 10.

(c) 5 × 13, 5 × 20

  • Solution: HCF = 5 × HCF(13, 20) = 5 × 1 = 5.

(d) 12 × 16, 12 × 20

  • Solution: HCF = 12 × HCF(16, 20) = 12 × 4 = 48.

In which of these cases is the HCF the same as the common multiplier?

  • Solution: In case (b), the HCF is 10, which is the common multiplier. This happens when the other two numbers (38 and 21) are co-prime (HCF = 1).

Efficient Procedures for HCF and LCM

Division Method for HCF and LCM

  • This method is similar to prime factorisation.
  • At each step, divide both numbers by a common prime factor.
  • Write the two quotients in the next row.
  • Continue till you get two numbers that do not have any common prime factors.

Example: Find the HCF of 84 and 180.

2 | 84, 180
2 | 42, 90
3 | 21, 45
  | 7, 15
  • HCF = 2 × 2 × 3 = 12.

Example: Find the LCM of 300 and 150.

2 | 300, 150
5 | 150, 75
5 | 30, 15
3 | 6, 3
  | 2, 1
  • HCF = 2 × 5 × 5 × 3 = 150.
  • LCM = 2 × 5 × 5 × 3 × 2 × 1 = 300.

Example: Find the HCF and LCM of 630 and 770.

2 | 630, 770
5 | 315, 385
7 | 63, 77
  | 9, 11
  • HCF = 2 × 5 × 7 = 70.
  • LCM = 2 × 5 × 7 × 9 × 11 = 6930.

Guna’s Faster Method

  • Guna says: “For the numbers 300 and 150, I can first directly divide both numbers by 50.”
50 | 300, 150
3  | 6, 3
   | 2, 1
  • HCF = 50 × 3 = 150.
  • LCM = 50 × 3 × 2 × 1 = 300.

Why this works: We need not restrict ourselves to dividing only one prime factor at a time. Both numbers can be divided by whatever common factor we are able to identify.

Question: Try this method for these pairs of numbers:

(a) 90 and 150

Solution:

30 | 90, 150
   | 3, 5
  • HCF = 30.
  • LCM = 30 × 3 × 5 = 450.

(b) 84 and 132

Solution:

12 | 84, 132
   | 7, 11
  • HCF = 12.
  • LCM = 12 × 7 × 11 = 924.

Property Involving both the HCF and the LCM

Question: Which is greater — the LCM of two numbers or their product?

Solution:

  • The LCM is never greater than the product of the two numbers.
  • This is because the product itself is a common multiple of the two numbers.

Relationship Between HCF, LCM, and Product

Example: Consider the numbers 105 and 95. Find their LCM.

  • Prime factorisation:
    • 105 = 3 × 5 × 7
    • 95 = 5 × 19
  • LCM = 3 × 5 × 7 × 19 = 1995.
  • Product in factorised form:
    • 105 × 95 = (3 × 5 × 7) × (5 × 19) = 3 × 5 × 5 × 7 × 19.

Is the LCM a factor of the product?

  • Yes, because LCM = 3 × 5 × 7 × 19, and product = 3 × 5 × 5 × 7 × 19.
  • Product = LCM × 5.
  • Notice that 5 is the HCF of 105 and 95.

General Property:

HCF × LCM = Product of the two numbers

Why does this happen?

  • Among the prime factors of two numbers, some are common to both factorisations, and the rest occur in only one of them.
  • The HCF contains all the common prime factors (with minimum occurrences).
  • The LCM contains all the prime factors (with maximum occurrences).
  • When we multiply HCF × LCM, we get all prime factors with their exact occurrences from both numbers, which equals the product.

Question: Explore whether the LCM is a factor of the product in the following cases. If yes, identify the number that the LCM should be multiplied by to get the product:

(a) 45, 105

Solution:

  • 45 = 3² × 5
  • 105 = 3 × 5 × 7
  • HCF = 3 × 5 = 15
  • LCM = 3² × 5 × 7 = 315
  • Product = 45 × 105 = 4725
  • LCM × HCF = 315 × 15 = 4725 ✓

(b) 275, 352

Solution:

  • 275 = 5² × 11
  • 352 = 2⁵ × 11
  • HCF = 11
  • LCM = 2⁵ × 5² × 11 = 8800
  • Product = 275 × 352 = 96800
  • LCM × HCF = 8800 × 11 = 96800 ✓

(c) 222, 370

Solution:

  • 222 = 2 × 3 × 37
  • 370 = 2 × 5 × 37
  • HCF = 2 × 37 = 74
  • LCM = 2 × 3 × 5 × 37 = 1110
  • Product = 222 × 370 = 82140
  • LCM × HCF = 1110 × 74 = 82140 ✓

Question: Explore whether this property holds when 3 numbers are considered.

Solution:

  • For three numbers, the property does not hold in the same way.
  • HCF × LCM ≠ Product of the three numbers (in general).

Figure it Out

Question 1: In the two rows below, colours repeat as shown. When will the blue stars meet next?

  • Row 1: Pattern repeats every 6 units.
  • Row 2: Pattern repeats every 8 units.
  • Solution: Blue stars will meet next at LCM(6, 8) = 24 units.

Question 2:

(a) Is 5 × 7 × 11 × 11 a multiple of 5 × 7 × 7 × 11 × 2?

Solution:

  • 5 × 7 × 11 × 11 = 5 × 7 × 11²
  • 5 × 7 × 7 × 11 × 2 = 2 × 5 × 7² × 11
  • For the first to be a multiple of the second, it must contain all prime factors of the second.
  • First number doesn’t have 2 or 7², so it is not a multiple.

(b) Is 5 × 7 × 11 × 11 a factor of 5 × 7 × 7 × 11 × 2?

Solution:

  • For the first to be a factor of the second, all its prime factors must be in the second.
  • First has 5, 7, 11². Second has 2, 5, 7², 11.
  • First has 11² but second has only 11, so it is not a factor.

Question 3: Find the HCF and LCM of the following (state your answers in the form of prime factorisations):

(a) 3 × 3 × 5 × 7 × 7 and 12 × 7 × 11

Solution:

  • First number: 3² × 5 × 7²
  • Second number: 12 × 7 × 11 = (2² × 3) × 7 × 11 = 2² × 3 × 7 × 11
  • Common primes: 3, 7.
  • HCF = 3 × 7 = 21.
  • All primes: 2, 3, 5, 7, 11.
  • LCM = 2² × 3² × 5 × 7² × 11 = 4 × 9 × 5 × 49 × 11 = 97020.

(b) 45 and 36

Solution:

  • 45 = 3² × 5
  • 36 = 2² × 3²
  • Common prime: 3.
  • HCF = 3² = 9.
  • All primes: 2, 3, 5.
  • LCM = 2² × 3² × 5 = 4 × 9 × 5 = 180.

Question 4: Find two numbers whose HCF is 1 and LCM is 66.

Solution:

  • 66 = 2 × 3 × 11.
  • Numbers with HCF = 1 are co-prime.
  • Possible pairs: (2, 33), (3, 22), (6, 11).
  • Verify: LCM(2, 33) = 66 ✓, LCM(3, 22) = 66 ✓, LCM(6, 11) = 66 ✓.

Question 5: A cowherd took all his cows to graze in the fields. The cows came to a crossing with 3 gates. An equal number of cows passed through each gate. Later at another crossing with 5 gates, again an equal number passed through each gate. The same happened at the third crossing with 7 gates. If the cowherd had less than 200 cows, how many cows did he have?

Solution:

  • Total number of cows must be divisible by 3, 5, and 7.
  • LCM(3, 5, 7) = 105.
  • Multiples of 105 less than 200: 105.
  • Answer: 105 cows.

Question 6: The length, width, and height of a box are 12 cm, 18 cm, and 36 cm respectively. Which of the following sized cubes can be packed in this box without leaving gaps?

(a) 9 cm

  • Solution: 9 is not a factor of 12, so No.

(b) 6 cm

  • Solution: 6 is a factor of 12, 18, and 36, so Yes.

(c) 4 cm

  • Solution: 4 is not a factor of 18, so No.

(d) 3 cm

  • Solution: 3 is a factor of 12, 18, and 36, so Yes.

(e) 2 cm

  • Solution: 2 is a factor of 12, 18, and 36, so Yes.

Question 7: Among the numbers below, which is the largest number that perfectly divides both 306 and 36?

(a) 36

  • Solution: 36 divides 36 but not 306 (306 ÷ 36 = 8.5).

(b) 612

  • Solution: 612 is greater than both numbers, so No.

(c) 18

  • Solution: 18 divides both 306 and 36. HCF(306, 36) = 18 ✓.

(d) 3

  • Solution: 3 divides both, but not the largest.

(e) 2

  • Solution: 2 divides both, but not the largest.

(f) 360

  • Solution: 360 is greater than both, so No.

Answer: (c) 18.

Question 8: Find the smallest number that is divisible by 3, 4, 5 and 7, but leaves a remainder of 10 when divided by 11.

Solution:

  • LCM(3, 4, 5, 7) = 420.
  • Numbers divisible by 420: 420, 840, 1260, …
  • Check remainder when divided by 11:
    • 420 ÷ 11 = 38 remainder 2.
    • 840 ÷ 11 = 76 remainder 4.
    • 1260 ÷ 11 = 114 remainder 6.
    • 1680 ÷ 11 = 152 remainder 8.
    • 2100 ÷ 11 = 190 remainder 10 ✓.
  • Answer: 2100.

Question 9: Children are playing ‘Fire in the Mountain’. When the number 6 was called out, no one got out. When the number 9 was called out, no one got out. But when the number 10 was called out, some people got out. How many children could have been playing initially?

(a) 72

  • Solution: 72 is divisible by 6, 9, and not by 10 ✓.

(b) 90

  • Solution: 90 is divisible by 6, 9, and also by 10, so some would get out ✗.

(c) 45

  • Solution: 45 is divisible by 9 but not by 6 ✗.

(d) 3

  • Solution: 3 is not divisible by 6 or 9 ✗.

(e) 36

  • Solution: 36 is divisible by 6 and 9 but not by 10 ✓.

(f) None of these

Answer: (a) 72 or (e) 36.

Question 10: Tick the correct statement(s). The LCM of two different prime numbers (m, n) can be:

(a) Less than both numbers

  • Solution: No, LCM of two primes is their product, which is greater than both.

(b) In between the two numbers

  • Solution: No, LCM = m × n, which is greater than both.

(c) Greater than both numbers

  • Solution: Yes ✓.

(d) Less than m × n

  • Solution: No, LCM = m × n.

(e) Greater than m × n

  • Solution: No, LCM = m × n.

Answer: (c).

Question 11: A dog is chasing a rabbit that has a head start of 150 feet. The dog jumps 9 feet every time the rabbit jumps 7 feet. In how many leaps does the dog catch up with the rabbit?

Solution:

  • Relative speed = Dog gains 9 – 7 = 2 feet per leap.
  • Distance to cover = 150 feet.
  • Number of leaps = 150 ÷ 2 = 75 leaps.

Question 12: What is the smallest number that is a multiple of 1, 2, 3, 4, 5, 6, 8, 9, 10?

Solution:

  • LCM(1, 2, 3, 4, 5, 6, 8, 9, 10).
  • Prime factorisations:
    • 1 = 1
    • 2 = 2
    • 3 = 3
    • 4 = 2²
    • 5 = 5
    • 6 = 2 × 3
    • 8 = 2³
    • 9 = 3²
    • 10 = 2 × 5
  • LCM = 2³ × 3² × 5 = 8 × 9 × 5 = 360.

Question 13: Here is a problem posed by the ancient Indian Mathematician Mahaviracharya (850 C.E.). Add together 8/15, 17/20, 11/36, and 1/6. What do you get? How can we find this sum efficiently?

Solution:

  • To add fractions, find the LCM of denominators: LCM(15, 20, 36, 6).
  • 15 = 3 × 5
  • 20 = 2² × 5
  • 36 = 2² × 3²
  • 6 = 2 × 3
  • LCM = 2² × 3² × 5 = 4 × 9 × 5 = 180.
  • Convert to common denominator:
    • 8/15 = 96/180
    • 17/20 = 153/180
    • 11/36 = 55/180
    • 1/6 = 30/180
  • Sum = (96 + 153 + 55 + 30) / 180 = 334/180 = 167/90.

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