ICSE Class 8 Maths Chapter 03 HCF And LCM

Chapter 3

H.C.F. And L.C.M.

Highest Common Factor (H.C.F.)

Factor

When a number divides another number completely (i.e. without leaving any remainder), it is called a factor of the second number.

e.g. 3 divides 18 exactly, leaving no remainder, so 3 is a factor of 18.

Similarly, other factors of 18 are: 1, 2, 6, 9 and 18 as each of these also divides 18 exactly.

Factors of 18 = 1, 2, 3, 6, 9 and 18.

Common Factor

A common factor of two or more given numbers is a number which divides each given number completely.

Since, factors of 18 = 1, 2, 3, 6, 9 and 18

and, factors of 24 = 1, 2, 3, 4, 6, 8, 12 and 24

Common factors of 18 and 24 = 1, 2, 3 and 6.

(Each of these common factors divides the given numbers 18 and 24 completely)

H.C.F.

The highest common factor (H.C.F.) of two or more given numbers is the greatest number (factor) which divides each of the given numbers completely.

As discussed above, the common factors of 18 and 24 are 1, 2, 3 and 6. Out of these common factors 6 is the greatest.

Highest Common Factor of 18 and 24 = 6

i.e. H.C.F. of 18 and 24 = 6

Test Yourself

1. Factors of 16 = ..., ..., ..., ... and ...

2. Factors of 20 = ..., ..., ..., ..., ... and ...

3. Factors of 28 = ..., ..., ..., ..., ... and ...

4. Common factors of 16, 20 and 28 are : ..., ... and ...

H.C.F. of 16, 20 and 28 = .........

Teacher's Note

Understanding factors and common factors helps us divide things fairly in real life, like splitting a pizza into equal slices or organizing items into equal groups.

Example 1

Find the H.C.F. of 36 and 84.

Solution

First Method (Factor method)

Factors of 36 = 1, 2, 3, 4, 6, 9, 12, 18 and 36

Factors of 84 = 1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42 and 84

Common factors of 36 and 84 = 1, 2, 3, 4, 6, 12

Highest Common Factors (i.e. H.C.F.) = 12 (Ans.)

Second Method (Prime factor method)

Since, 36 = 2 × 2 × 3 × 3

and, 84 = 2 × 2 × 3 × 7

H.C.F. of 36 and 84 = Product of common prime factors

= 2 × 2 × 3 = 12 (Ans.)

Teacher's Note

Prime factorization is like breaking down a number into its smallest building blocks, similar to how a recipe breaks down into individual ingredients.

Third Method (Division method)

36 / 84 with quotient 2 and remainder 72

12 / 36 with quotient 3 and remainder 36

H.C.F. of 36 and 84 = Last divisor = 12 (Ans.)

To Find The H.C.F. Of Three Or More Given Numbers By Division Method

Steps: (i) First of all, find the H.C.F. of any two of the given numbers.

(ii) Now, find the H.C.F. of the H.C.F. (obtained in step (i)) and the third number. Continue till the last H.C.F. is obtained.

(iii) The last H.C.F. is the H.C.F. of all the given numbers.

Teacher's Note

The division method is efficient for finding H.C.F. because it systematically reduces numbers by division, much like how a calculator simplifies fractions.

Example 2

Find (using division method) the H.C.F. of:

(i) 306, 234 and 405 (ii) 525, 420, 1245 and 1080.

Solution

(i) Step 1

Find the H.C.F. of 306 and 234.

234 / 306 with quotient 1 and remainder 234

72 / 234 with quotient 3 and remainder 216

18 / 72 with quotient 4 and remainder 72

H.C.F. of 306 and 234 = 18

Step 2

Find the H.C.F. of 18 and 405

18 / 405 with quotient 22 and remainder 36

45 (calculated)

36 (calculated)

9 / 18 with quotient 2 and remainder 18

Hence, H.C.F. of the given numbers = 9 (Ans.)

(ii) Step 1

Find the H.C.F. of 525 and 420.

420 / 525 with quotient 1 and remainder 420

105 / 420 with quotient 4 and remainder 420

H.C.F. of 525 and 420 = 105

Step 2

Find the H.C.F. of 105 and 1245.

105 / 1245 with quotient 11 and remainder 105

195 (calculated)

105 (calculated)

90 / 105 with quotient 1 and remainder 90

15 / 90 with quotient 6 and remainder 90

H.C.F. of 105 and 1245 = 15

Step 3

Finally, find the H.C.F. of 15 and 1080.

15 / 1080 with quotient 72 and remainder 105

30 (calculated)

30 (calculated)

H.C.F. of 15 and 1080 = 15

H.C.F. of 525, 420, 1245 and 1080 = 15 (Ans.)

Teacher's Note

Finding H.C.F. of multiple numbers step-by-step is like finding the common ground between multiple people in a group discussion.

Example 3

Find the greatest number that will divide 185 and 300 leaving remainders 3 and 6 respectively.

Solution

Since on dividing 185 by the required number, the remainder is 3.

185 - 3 = 182 will be exactly divisible by the required number.

Similarly, 300 - 6 = 294 will also be exactly divisible by the required number.

The required number is H.C.F. of 182 and 294.

182 = 2 × 7 × 13 and 294 = 2 × 3 × 7 × 7 (Resolving into prime factors)

Required greatest no. = 2 × 7 = 14 (Ans.)

Teacher's Note

This concept applies to real-world scenarios like arranging items on shelves with specific gaps or organizing schedules with overlaps.

If two numbers are co-prime, i.e., they do not have any common divisor except one; their H.C.F. = 1

e.g. (i) H.C.F. of 16 and 25 is 1

(ii) H.C.F. of 9 and 80 is 1 and so on.

Lowest Common Multiple (L.C.M.)

Multiple

When a number divides another number completely, the second number is called a multiple of the first number.

e.g. 3 divides 18 completely, so 18 is a multiple of 3.

Multiples of 3 = 3, 6, 9, 12, 15, 18, 21, .....................

Multiples of 5 = 5, 10, 15, 20, 25, ........... and so on.

Common Multiple

A number, which can be divided completely by the given two or more numbers, is called their common multiple.

Since, multiples of 4 = 4, 8, 12, 16, 20, 24, 28, 32, 36, 40, ......

multiples of 6 = 6, 12, 18, 24, 30, 36, ........

and, multiples of 12 = 12, 24, 36, 48, ............

Common multiples of 4, 6 and 12 = 12, 24, 36, ......

(Each of these common multiples is completely divisible by the given numbers 4, 6 and 12).

L.C.M.

It is the lowest number which is completely divisible by each of the given numbers.

Since, the common multiples of 4, 6 and 12 = 12, 24, 36, .....

The lowest common multiple of 4, 6 and 12 is 12.

And, 12 is the smallest number which is completely divisible by each of the given numbers 4, 6 and 12.

Teacher's Note

L.C.M. is useful in scheduling problems, like finding when two events will happen simultaneously again, such as when two buses with different schedules arrive at the same stop together.

Test Yourself

5. Multiples of 3 = ..., ..., ..., ..., ..., ..., ..., ..., ..., ..., ...........................

multiples of 4 = ..., ..., ..., ..., ..., ..., ..., ...........................

and, multiples of 6 = ..., ..., ..., ..., ..., ..., ...........................

Common multiples of 3, 4 and 6 = ..., ..., ................................

and, L.C.M. of 3, 4 and 6 = .....

Example 4

Find L.C.M. of 24, 36 and 40.

Solution

First Method (Index method)

24 = 2 × 2 × 2 × 3 = 2^3 × 3

36 = 2 × 2 × 3 × 3 = 2^2 × 3^2

40 = 2 × 2 × 2 × 5 = 2^3 × 5

L.C.M. = Product of each prime factor with highest index used.

= 2^3 × 3^2 × 5

= 8 × 9 × 5 = 360 (Ans.)

Second Method (Division method)

2 | 24, 36, 40

2 | 12, 18, 20

2 | 6, 9, 10

3 | 3, 9, 5

1, 3, 5

L.C.M. = Product of each divisor and quotients

= 2 × 2 × 2 × 3 × 1 × 3 × 5

= 360 (Ans.)

Teacher's Note

The index method is like organizing ingredients by their importance and quantity needed in a recipe - taking the highest amount of each ingredient available.

Example 5

Find the smallest number which when divided by 48 and 60 respectively, leaves a remainder 7 in each case.

Solution

Find the L.C.M. of the given numbers 48 and 60.

Since, 48 = 2 × 2 × 2 × 2 × 3 = 2^4 × 3

and, 60 = 2 × 2 × 3 × 5 = 2^2 × 3 × 5

L.C.M. = 2^4 × 3 × 5 = 240

L.C.M. of 48 and 60 is 240 means, 240 is the smallest number which is exactly divisible by 48 and 60.

So, the required smallest number, which on dividing by 48 and 60, will leave remainder 7 in each case

= 240 + 7 = 247 (Ans.)

Example 6

What is the smallest number which when increased by 3 is divisible by 27, 35, 25 and 21.

Solution

The smallest number which is divisible by 27, 35, 25 and 21 is the L.C.M. of 27, 35, 25 and 21.

The required number = L.C.M. of the given numbers minus 3.

Since, 27 = 3 × 3 × 3 = 3^3; 35 = 5 × 7;

25 = 5 × 5 = 5^2 and 21 = 3 × 7

L.C.M. = 3^3 × 5^2 × 7 = 4725

And, the required no. = 4725 - 3 = 4722 (Ans.)

Teacher's Note

This type of problem mirrors real scheduling situations, like finding when multiple repeating events will next occur together, then adjusting for an offset.

1. For any two co-prime numbers, their L.C.M. is equal to their product.

e.g. (i) L.C.M. of 8 and 15 = 8 × 15 = 120

(ii) L.C.M. of 12 and 35 = 12 × 35 = 420 and so on.

2. For any two numbers:

The product of their L.C.M. and H.C.F. = The product of the numbers.

Example 7

The L.C.M. of two numbers is 2079 and their H.C.F. is 27. If one of the numbers is 189, find the other.

Solution

For any two numbers;

One no. × other no. = Their L.C.M. × their H.C.F.

=

189 × other no. = 2079 × 27

=

The other no. = (2079 × 27) / 189 = 297 (Ans.)

Example 8

Two persons take steps of 80 cm and 90 cm respectively. If they start in step, how far will they walk before they are in step again?

Solution

The required distance = The L.C.M. of 80 cm and 90 cm

| 80 = 2^4 × 5

= 2^4 × 3^2 × 5 cm

| 90 = 2 × 3^2 × 5

= 720 cm (Ans.)

Example 9

Find the smallest number which when divided by 35, 45 and 55 leaves the remainders 17, 27 and 37 respectively.

Solution

For divisor 35, remainder = 17 and 35 - 17 = 18.

For divisor 45, remainder = 27 and 45 - 27 = 18

For divisor 55, remainder = 37 and 55 - 37 = 18

Every pair of divisor and remainder has same difference i.e. 18

Hence, the required number = L.C.M. of 35, 45 and 55 minus 18.

Since, 35 = 5 × 7; 45 = 3 × 3 × 5 and 55 = 5 × 11

L.C.M. of 35, 45 and 55 = 3 × 3 × 5 × 7 × 11

= 3465

The required number = 3465 - 18 = 3447 (Ans.)

Teacher's Note

This problem demonstrates how finding patterns in remainders helps solve complex scheduling and distribution problems in real-world logistics.

Exercise 3 (A)

1. Find the H.C.F. of:

(i) 540, 720 and 450

(ii) 805, 1127 and 1449

(iii) 1836, 810, 1296 and 702.

2. What is the greatest number which when divides 398 and 436, leaves 7 and 11 respectively as remainders.

3. Find the greatest number which divides 1750 and 2000 leaving 48 and 2 respectively as remainders.

4. Find L.C.M. of:

(i) 72, 84 and 126

(ii) 294, 420 and 504

(iii) 24, 36, 108 and 192

5. Find the least number which when divided by 36, 48 and 112; leaves no remainder.

6. Find the smallest number which when divided by 12, 20, 30 and 60; leaves a remainder 5 each time.

7. What is the least number which when increased by 3 is exactly divisible by 27, 35 and 21?

8. (i) Find the least number that can be divided exactly by all the even numbers between 10 and 20.

(ii) Find the least number that can be divided exactly by all odd numbers between 20 and 30.

9. What is the least number which when decreased by 5 is divisible by 36, 48, 21 and 28?

10. The H.C.F. of two numbers is 119 and their L.C.M. is 11781. If one of the numbers is 1071, find the other.

11. The product of two numbers is 20736 and their H.C.F. is 24. Find their L.C.M.

12. Two persons take steps of 64 cm and 84 cm respectively. If they start in step, how far will they walk before they are in step again?

13. Four bells are ringing at intervals of 12, 16, 24 and 36 minutes. They start ringing simultaneously at 12 O'clock. Find when will they again ring together?

14. Show that 280 and 297 are prime to each other.

Show that the H.C.F. of the given numbers is 1.

15. A farmer has 2494 sheep and 2193 lambs. He farms them into flocks, keeping sheep and lambs separate and having the same number of animals in each flock. If these flocks are as large as possible, find:

(i) the maximum number of animals in each flock and

(ii) total number of flocks required for the purpose.

16. Find the smallest number which when divided by 42 and 54 leaves the remainders 34 and 46 respectively

17. Find the smallest number which when divided by 24, 36 and 48 leaves the remainders 21, 33 and 45 respectively.

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