The LCM and HCF concepts are very useful in solving problems involving fractions, simplifying algebraic expressions, and finding solutions to equations. Some questions based on LCM and HCF concepts are as follows:

- Find the LCM and HCF of two or more numbers
- Simplification of fractions
- Find the smallest or largest number satisfying given conditions
- Calculate the time taken by two or more persons to complete a task working together.
- Total number of rotations or revolutions required by two or more wheels to come back to their initial position
- Find the least n-digit number which when divided by the given numbers leaves the same or different remainder.
- How many pairs of numbers are there whose product is given
- Arrangements in rows and columns, etc.

**Introduction to LCM and HCF**

The concepts of LCM (Least Common Multiple) and HCF (Highest Common Factor) are commonly used in mathematics, particularly in the study of integers. (LCM and HCF concepts)

The LCM of two or more integers is the smallest positive integer that is a multiple of each of the given integers. On the other hand, the HCF of two or more integers is the largest positive integer that divides each of the given integers without leaving a remainder. HCF is also known as GCD (Greatest Common Divisor).

**Multiple**

The multiple of a number is always exactly divisible by the given number. (LCM and HCF concepts)

Ex: The multiple of 6 are: 6, 12, 18, 24,â€¦ etc.

**Factor or Divisor**

A factor or divisor always divides the given number exactly i.e. leaving no remainder. (LCM and HCF concepts)

Ex: The factors of 6 are: 1, 2, 3, 6

Note: Multiples of a number are unlimited in number but divisors are limited.

Let two numbers be x and y. Their HCF is â€˜Hâ€™ and LCM is â€˜Lâ€™.

HCF â‰¤ x and y

LCM â‰¥ x and y

Here, H is the largest number that can divide both x and y.

x= H.a

y= H.b

Here a & b are constants. (LCM and HCF concepts)

LCM (L)= H.a.b

So, LCMÃ—HCF=First number Ã—Second number

Note: This formula does not apply to more than two numbers.

**Prime Factor**

A prime number, that divides a given number without leaving a remainder.

**Prime Factorization**

It is the process of expressing a positive integer as a product of its prime factors.

Example: 420= 2^{2}Ã—3Ã—5Ã—7

**How to find LCM**

The prime factorization method is the best way to find out the LCM of given numbers.

Example: Find out the LCM of 240, 420, and 675

**Step 1: Prime Factorization of each number.**

240= 2^{4}Ã—3Ã—5

420= 2^{2}Ã—3Ã—5Ã—7

675= 3^{3}Ã—5^{2}

**Step 2: Write all the prime numbers with their highest exponent and multiply them.**

LCM= 2^{4}Ã—3^{3}Ã—5^{2}Ã—7= 75600

This means that there is no number below 75600 which is divisible by 240, 420, and 675.

**How to find HCF**

The prime factorization method is the best way to find out the HCF of given numbers.

Example: Find out the HCF of 240, 420, and 675

**Step 1: Prime Factorization of each number.**

240= 2^{4}Ã—3Ã—5

420= 2^{2}Ã—3Ã—5Ã—7

675= 3^{3}Ã—5^{2}

**Step 2: Write all the prime numbers which are common in all, with the respective exponents.**

HCF= 3Ã—5= 15

This means that there is no number greater than 15 that can divide 240, 420, and 675.

**Difference between LCM and HCF**

LCM | HCF |

Least Common Multiple | Highest common Factor |

The smallest positive integer that is divisible by two or more numbers | The largest positive integer that can divide, two or more numbers |

LCM of two or more numbers is always greater than or equal to each of the given numbers | HCF of two or more numbers is always less than or equal to each of the given numbers |

The LCM of two co-prime numbers is equal to the product of those numbers | The HCF of two co-prime numbers is equal to 1 |

**Co-prime Numbers: **Two numbers that have no common divisor other than 1 are called Coprime or Relatively Prime numbers.

Example: 9 and 20

LCM= 9Ã—20=180

HCF= 1

**Shortcut for finding HCF**

Let x, y, and z be positive integers whose HCF is H. Then

x=H.a

y=H.b

z=H.c

Here a, b, and c are co-prime numbers. (LCM and HCF Concepts)

Now, HCF (x-y, y-z, z-x)= H

**Example: Find out the HCF of 240, 420, and 675**

Step 1: 420-240= 180, 675-420=255, 675-240= 435

Step 2: Repeat this process again for 180, 255, and 435

255-180= 75

435-255= 180

435- 180= 255

Now, we can easily find the HCF of these smaller numbers.

HCF= 15

Note: We will use this method in case of large numbers to reduce the complexity of the calculation. (LCM and HCF Concepts)

**LCM and HCF of fractions**

HCF of two or more fractions= \frac{HCF\;of\;Numerators}{LCM\;of\;Denominators}.

LCM of two or more fractions= \frac{LCM\;of\;Numerators}{HCF\;of\;Denominators}.

**Que:** Find the LCM and HCF of \frac{15}{16},\frac{7}{8},\frac{25}{36}.

**Solution: **LCM of Numerators= 3Ã—7Ã—5^{2}= 525

15=3Ã—5

7=7

25=5^{2}

HCF of Denominators= 2^{2}=4

16= 2^{4}

8= 2^{3}

36= 2^{2}Ã—3^{2}

LCM of the fractions= \frac{525}{4} â€¦â€¦â€¦.(i)

HCF of Numerators= 1

LCM of Denominators= 2^{4}Ã—3^{2}= 144

HCF of the fractions= \frac{1}{144} â€¦â€¦â€¦.(ii)

**Practice Exercise**

**Que 1: **The LCM and HCF of the two numbers are 936 and 4 respectively. If one number is 72 then the other number is

(a) 52

(b) 62

(c) 44

(d) 48

**Solution: **936Ã—4=72Ã—k

k=52 (LCM and HCF Concepts)

Option (a) is correct

**Que 2: **At 10 in the morning, 4 bells ring together. If they ring after 7, 8, 11, and 12 seconds respectively, how many times will they ring together again in the next 12 hours?

(a) 18

(b) 21

(c) 23

(d) 26

**Solution:** LCM(7,8,11,12)= 1848

Number of seconds in 12 hours= 12Ã—60Ã—60= 43200 seconds

The quotient of the division (43200/1848)= 23

Option (c) is correct

**Que 3: **If the HCF and LCM of some numbers like 2472, 1248, and N, etc. are 12 and 2^{3}Ã—3^{2}Ã—5Ã—103Ã—107 respectively, then what can be the value of N?

(a) 2^{2}Ã—3^{2}Ã—7

(b) 2^{2}Ã—3^{3}Ã—103

(c) 2^{2}Ã—3^{2}Ã—5

(d) None of these

**Solution: **N=12k, Here k is a positive integer

Only option (c) **can be** a possible value of N.

Option (a)â‡’ 7 is present, hence disqualified

Option (b)â‡’ 3^{3} is present, hence disqualified (LCM and HCF Concepts)

Option (c) is correct

**Que 4: **What is the minimum number of square tiles required to pave the floor of a room 15 m 17 cm long and 9 m 2 cm broad?

(a) 764

(b) 814

(c) 996

(d) 698

**Solution: **15 m 17 cm= 1517 cm

9 m 2 cm= 902 cm

The side length of square tile= HCF(1517, 902)= 41

Number of paved tiles along the length of the room= 1517Ã·41= 37

Number of paved tiles along the width of the room= 902Ã·41= 22

The number of tiles required= 37Ã—22= 814 (LCM and HCF Concepts)

Option (b) is correct

**Que 5: **Three runners start running around a circular track at 5 am. They can complete one round in 2, 4, and 5.5 minutes respectively. When will they meet at the starting point?

(a) 10:44 AM

(b) 10:24 AM

(c) 11:00 AM

(d) 11:20 AM

**Solution:** LCM (2, 4, 11/2)= 44

It means that after 44 minutes all will be at the starting point together. (LCM and HCF Concepts)

Option (a) is correct

**Que 6: **The H.C.F. of the two numbers is 27 and the two factors of their LCM are 12 and 15. What is the largest number?

(a) 810

(b) 540

(c) 405

(d) 135

**Solution:** HCF= 27 (LCM and HCF Concepts)

Let the numbers be 27x and 27y

LCM= 27xy= 3^{3}.xy

12=2^{2}Ã—3

15= 3Ã—5

LCM= 3^{3}Ã—2^{2}Ã—5

(x,y)={(2^{2}Ã—5, 1), (2^{2},5)}

Larger number= 27Ã—2^{2}Ã—5=540

Option (b) is correct.

LCM and HCF Problems: Click Here