HCF and LCM - Solved Questions and Answers

HCF (Highest Common Factor) is the largest number that divides two or more numbers without leaving a remainder and LCM (Least Common Multiple) is the smallest positive number that is a multiple of two or more given numbers.

HCF and LCM questions and answers are provided below for you to learn and practice.

Question 1: Find the greatest number that will divide 72, 96, and 120, leaving the same remainder in each case.

Solution:

To find the greatest number that will divide these numbers leaving the same remainder, we need to find the Highest Common Factor (HCF) of the differences between the numbers. The differences are as follows:

96 - 72 = 24
120 - 96 = 24

HCF(24, 24) = 24

Therefore, the greatest number that will divide 72, 96, and 120 leaving the same remainder in each case is 24.

Question 2: If the HCF of two numbers is 12 and their LCM is 360, find the numbers.

Solution:

We know that the product of the HCF and LCM of two numbers is equal to the product of the two numbers.

So, for two numbers a and b with HCF = 12 and LCM = 360:

HCF × LCM = a × b
12 × 360 = a × b
4320 = a × b

Now, we need to find two numbers whose product is 4320 and HCF is 12. There can be multiple pairs of numbers that satisfy this condition, and one such pair is:

a = 120 and b = 36

Because 120 × 36 = 4320 and the HCF of 120 and 36 is 12.

Question 3: Find the HCF of 36, 48, and 72.

Solution:

• Prime factorization of 36: 2² × 3²
• Prime factorization of 48: 2⁴ × 3
• Prime factorization of 72: 2³ × 3²

Common factors: 2² and 3 (take the minimum power)

So, the HCF of 36, 48, and 72 is 2² × 3 = 12.

Question 4: What is the largest three-digit number that is exactly divisible by the HCF of 24 and 36?

Solution:

The HCF of 24 and 36 is 12. To find the largest three-digit number that is exactly divisible by 12, we need to find the largest multiple of 12 that is less than 1000.

The largest multiple of 12 less than 1000 is 996 (83 × 12 = 996).

So, the largest three-digit number exactly divisible by the HCF of 24 and 36 is 996.

Question 5: The sum of two numbers is 1001, and their HCF is 7. Find the numbers.

Solution:

a + b = 1001 (Sum of the two numbers)
HCF(a, b) = 7

We know that if the HCF of two numbers divides their sum, then it also divides the difference of the two numbers.

So, a - b is divisible by 7.

Now, we look for pairs of numbers whose difference is divisible by 7 and whose sum is 1001.

One such pair is a = 504 and b = 497.

504 - 497 = 7 (divisible by the HCF)
504 + 497 = 1001

Therefore, the two numbers are 504 and 497.

Question 6: Find the HCF of 32 and 14 by Listing Factors Method.

Solution:

First, list down the factors of 32 and 14. 

• The factors of 32 are: 1, 2, 4, 8, 16, 32
• The factors of 14 are: 1, 2, 7, 14

We can see that 1, 2 are the only common factors of 32 and 14. Whereas 2 is the greatest among all the common factors.
Hence, HCF of 32 and 14 is 2.

Question 7: Find the HCF of 80 and 90 by Prime Factorization

Solution:

• The prime factors of 80: 2 * 2 * 2 * 2 * 5; 
• The prime factors of 90: 2 * 3 * 3 * 5.

We can see that 2, 5 are the only common factors of 80 and 90, Now, the HCF of 80 and 90 will be the product of the common prime factors, which are 2 and 5. 
Hence, HCF of 80 and 90 is 10.

Question 8: Find the HCF of 30 and 42 by Division Method

Solution:

To find the HCF of 30 and 42, we can use the above-mentioned algorithm, which can be used as follows:

Hence, the HCF of 30 and 42 is 6.

Solved Questions on LCM

Question 1: Find the LCM of 12, 18, and 24.

Solution:

• 12 = 2² × 3
• 18 = 2 × 3²
• 24 = 2³ × 3

Now, to find the LCM, we need to take the highest power of each prime factor that appears in any of the numbers:

• The highest power of 2 is 2³.
• The highest power of 3 is 3².

Multiplying these together gives us:

LCM = 2³ × 3² = 8 × 9 = 72

So, the LCM of 12, 18, and 24 is 72

Question 2: The LCM of the two numbers is 360, and their HCF is 24. If one of the numbers is 120, find the other number.

Solution:

We know that the product of the HCF and LCM of two numbers is equal to the product of the two numbers.

So, for two numbers a and b with HCF = 24 and LCM = 360:

HCF × LCM = a × b
24 × 360 = 120 × b [Given a = 120]
(24 × 360)/120 = b
24 × 3 = b
Thus, b = 72.

Question 3: A factory manufactures products in batches of 16, 24, and 32 units. What is the minimum number of units the factory needs to produce so that each batch can be formed exactly?

Solution:

To find the minimum number of units the factory needs to produce so that each batch size (16, 24, and 32) can be formed exactly, we need to find the least common multiple (LCM) of these batch sizes.

The prime factorization of each batch size is as follows:

• 16 = 2⁴
• 24 = 2³ × 3
• 32 = 2⁵

To find the LCM, we take the highest power of each prime factor that appears in any of the batch sizes:

• The highest power of 2 is 2⁵.
• The highest power of 3 is 3¹.

So, the LCM of 16, 24, and 32 is 2⁵ × 3¹ = 32 × 3 = 96.

Question 4: Find the LCM of two positive integers 2 and 6 by the Listing Method

Solution:

• Multiples of 2: 2, 4, 6, 8, 10, 12, 14...
• Multiples of 6: 6, 12, 18, 24, 30…

The common multipliers of 2 and 6 are 6, 12..., So, the least common multiple is 6.
Hence, LCM(2, 6) = 6

Question 5: Find the LCM of two positive integers 120 and 300 by Prime Factorization Method

Solution:

• The prime factorization of 120 are: 2*2*2*3*5 = 2³*3¹*5¹
• The prime factorization of 300 are: 2*2*3*5*5 = 2²*3¹*5²

Now, find the product of only those factors that have the highest powers among these. This will be, 2³ * 3¹ * 5² = 8 * 3 *  25 = 600

Hence, LCM(120, 300) = 600

Question 6: Let's take two positive integers 3 and 4, the task is to find the LCM(3, 4) by Prime Factorization Method.

Solution:

The LCM is the product of all these prime numbers.
Hence, LCM(3, 4) = 12

Miscellaneous Problems - HCF and LCM

Problem 1: Two numbers are in the ratio of 5:11. If their HCF is 7, find the numbers.

Solution:

Let the numbers be 5m and 11m. 

Since 5:11 is already the reduced ratio, 'm' has to be the HCF. 

So, the numbers are 5 x 7 = 35 and 11 x 7 = 77.  

Problem 2: Find the length of the plank which can be used to measure exactly the lengths 4 m 50 cm, 9 m 90 cm, and 16 m 20 cm in the least time.

Solution:

Let us first convert each length to cm. 

So, the lengths are 450 cm, 990 cm, and 1620 cm. 

Now, we need to find the length of the largest plank that can be used to measure these lengths as the largest plank will take the least time. 

For this, we need to take the HCF of 450, 990, and 1620. 

450 = 2 x 3 x 3 x 5 x 5 = 2 x 3² x 5² 
990 = 2 x 3 x 3 x 5 x 11 = 2 x 3² x 5 x 11 
1620 = 2 x 2 x 3 x 3 x 3 x 3 x 5 = 2² x 3⁴ x 5 

Therefore, HCF (450, 990, 1620) = 2 x 3 x 3 x 5 = 90 

Thus, we need a plank of length 90 cm to measure the given lengths in the least time.

Problem 3: Find the greatest number which on dividing 70 and 50 leaves remainders 1 and 4 respectively. 

Solution:

The required number leaves remainders 1 and 4 on dividing 70 and 50 respectively. 

This means that the number exactly divides 69 and 46. 

So, we need to find the HCF of 69 = 3 x 23 and 46 = 2 x 23. 

HCF (69, 46) = 23 Thus, 23 is the required number. 

Problem 4: Find the largest number that divides 64, 136, and 238 to leave the same remainder in each case. 

Solution:

To find the required number, we need to find the HCF of (136-64), (238-136), and (238-64), i.e., HCF (72, 102, 174). 

72 = 2³ x 3² 
102 = 2 x 3 x 17 
174 = 2 x 3 x 29 

Therefore, HCF (72, 102, 174) = 2 x 3 = 6 

Hence, 6 is the required number. 

Problem 5: Find the least number which when divided by 5,7,9 and 12, leaves the same remainder of 3 in each case.

Solution:

In these types of questions, we need to find the LCM of the divisors and add the common remainder (3) to it. 

So, LCM (5, 7, 9, 12) = 1260 

Therefore, required number = 1260 + 3 = 1263   

Problem 6: Find the largest four-digit number exactly divisible by 15,21 and 28. 

Solution:

The largest four-digit number is 9999. 

Now, LCM (15, 21, 28) = 420 On dividing 9999 by 420, we get 339 as the remainder. 

Therefore, the required number is 9999-339 = 9660   

Problem 7: Two numbers are in the ratio 2:3. If the product of their LCM and HCF is 294, find the numbers.

Solution:

Let the common ratio be 'm'. 
So, the numbers are 2m and 3m. 

Now, we know that the Product of numbers is = Product of LCM and HCF. 

⇒ 2m x 3m = 294 
⇒ m² = 49 
⇒ m = 7 

Therefore, the numbers are 14 and 21.

Problem 8: Find the least number which when divided by 6,7,8 leaves a remainder of 3, but when divided by 9 leaves no remainder. 

Solution:

LCM (6, 7, 8) = 168 So, the number is of the form 168m + 3. 

Now, 168m + 3 should be divisible by 9. 

We know that a number is divisible by 9 if the sum of its digits is a multiple of 9. 

For m = 1, the number is 168 + 3 = 171, the sum of whose digits is 9. 

Therefore, the required number is 171.   

Word Problems on HCF and LCM

Problem 1: The policemen at three different places on the ground blow a whistle after every 42 sec, 60 sec, and 78 sec respectively. If they all blow the whistle simultaneously at 9:30:00 hours, then at what time do they whistle again together?

Solution:

They all will whistle again at the same time after an interval that is equal to the LCM of their individual whistle-blowing cycles. 

So, LCM (42, 60, 78) = 2 x 3 x 7 x 10 x 13 = 5460 

Therefore, they will blow the whistle again simultaneously after 5460 sec, i.e., after 1 hour 31 minutes, i.e., at 11:01:00 hours.   

Problem 2: A rectangular field of dimension 180m x 105m is to be paved by identical square tiles. Find the size of each tile and the number of tiles required. 

Solution:

We need to find the size of a square tile such that a number of tiles cover the field exactly, leaving no area unpaved. 

For this, we find the HCF of the length and breadth of the field. 

HCF (180, 105) = 15 

Therefore, size of each tile = 15m x 15m 

Also, number of tiles = Area of field / Area of each tile 

⇒ Number of tiles = (180 x 105) / (15 x 15) 
⇒ Number of tiles = 84 

Hence, we need 84 tiles, each of size 15m x 15m. 

Problem 3: Three rectangular fields having areas of 60 m², 84 m^2, and 108 m² are to be divided into identical rectangular flower beds, each having a length of 6 m. Find the breadth of each flower bed. 

Solution:

We need to divide each large field into smaller flower beds such that the area of each bed is same. 

So, we find the HCF of the larger fields which gives us the area of the smaller field. HCF (60, 84, 108) = 12 

Now, this HCF is the area (in m²) of each flower bed. 

Also, area of a rectangular field = Length x Breadth 

⇒ 12 = 6 x Breadth 
⇒ Breadth = 2 m 

Hence, each flower bed would be 2 m wide. 

Problem 4: Find the maximum number of students among whom 182 chocolates and 247 candies can be distributed such that each student gets the same number of each. Also, find the number of chocolates and candies each student will get.

Solution:

We need to find the HCF of the number of chocolates and candies available, which would give us the number of students. 

HCF (182, 247) = 13 

So, there can be 13 students. 

Also, Number of chocolates for each student = 182 / 13 = 14 

Number of toffees for each student = 247 / 13 = 19

Problem 5: A bell rings every 8 minutes. A second bell rings every 12 minutes.  If all the two bells ring at the same time at 4 AM, at what other time will they all ring together?

Solution:

We need to find LCM of 8 and 12.

LCM of 8 and 12 will be 24 min.

The bells will ring together again at 4 AM+ 24 minutes = 4: 24 AM.

Greatest Common Divisor

Greatest Common Divisor

Lowest Common Multiple – LCM