# Problems on H.C.F and L.C.M – Aptitude Questions

HCF (Highest Common Factor) and LCM (Least Common Multiple) concepts are the foundation of many mathematical operations and are essential in solving complex problems. HCF and LCM problems challenge your ability to find the greatest common factor and the smallest common multiple of numbers, and they require both logical and mathematical skills. So get ready to exercise your brain as we delve into the world of HCF and LCM problems and explore the exciting ways they can be used to solve challenging aptitude questions!

## What is HCF (Highest Common Factor)?

The Highest Common Factor (HCF) of two numbers is the highest possible number that divides both numbers completely. The Highest Common Factor (HCF) is also known as the Greatest Common Divisor (GCD).

H.C.F. is also called the greatest common factor (GCF).

## How to Find HCF?

There are 3 methods to calculate the HCF of two numbers:

1. HCF by listing factors method
2. HCF by prime factorization
3. HCF by division method

### 1. HCF by Listing Factors Method

Here, we list the factors of each number and find the common factors of those numbers. Then, among the common factors, we determine the highest common factor.

#### Example:

Find the HCF of 32 and 14.

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.

### 2. HCF by Prime Factorization

We can find HCF using Prime factorization method of the given numbers.

Follow the steps below to calculate the HCF of given numbers using the prime factorization method:

1. First, Find the common prime factors of the given numbers.
2. Multiply these common prime factors to obtain the HCF of those numbers.

#### Example:

Find the HCF of 80 and 90.

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.

### 3. HCF by Division Method

The HCF of two numbers can be calculated using the division method.

Follow the steps below to calculate the HCF of given numbers using the Division Method

1. First, we divide the larger number by the smaller number and check the remainder.
2. Then, make the remainder of the previous step as the new divisor and the divisor of the previous step becomes the new dividend. After this we perform the long division again.
3. Continue the long division process till we get the remainder as 0. It should be noted that the last divisor will be the HCF of those two numbers.

#### Example:

Find the HCF of 30 and 42.

Solution: HCF of 30 and 42

Hence, HCF of 30 and 42 is 6.

## What is LCM (Least Common Multiple)?

In arithmetic, the LCM or least common multiple of two numbers a and b, is denoted as LCM (a,b) is the smallest or least positive integer that is divisible by both a and b

LCM is also called the Least Common Divisor

### Example:

Let’s take two positive integers 3 and 4, the task is to find the LCM(3, 4).
Solution:

• Multiples of 3: 3,6,9,12,15,18,21,24…
• Multiples of 4: 4,8,12,16,20,24, 28…

The common multipliers of 3 and 4 are 12, 24…, So, the least common multiple is 12.

Hence, LCM(3, 4) = 12

## How to Find LCM?

There are 3 methods to find the least common multiple of two numbers.

1. LCM by Listing Method
2. LCM by Prime Factorization Method
3. LCM using Division Method

### 1. LCM by Listing Method:

We can find out the common multiples of two or more numbers by listing their multiples. And out of these common multiples and the least common multiple is considered to be the LCM of two given numbers .

Follow the steps below to calculate the LCM of the two numbers A and B by the listing method:

1. First, list down first few multiples of A and B.
2. Mark the common multiples from the multiples of both numbers.
3. Select the smallest marked common multiple. Hence, this results in the LCM(A, B).

#### Example:

Find LCM of two positive integers 2 and 6.

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

### 2. LCM by Division Method:

We can find LCM using Division method of the given numbers. This can be done by dividing the numbers by a common prime number, and these prime factors are used to calculate the LCM of those numbers.

Follow the steps below to calculate the LCM of the two numbers A and B by Division Method:

1. First, find a prime number which is a factor of at least one of the given numbers. Write this prime number on the left of the given numbers.
2. If the prime number in step 1 is a factor of the number, then divide the number by the prime and write the quotient below it. If the prime number in step 1 is not a factor of the number, then write the number in the row below as it is. Continue the steps until 1 is left in the last row.

#### Example:

Let’s take two positive integers 3 and 4, the task is to find the LCM(3, 4).
Solution: LCM(3,4) = 12

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

### 3. LCM by Prime Factorization Method:

We can find LCM using Prime factorization method of the given numbers.

Follow the steps below to calculate the LCM of two numbers using the prime factorization method:

1. First, find the prime factors of the given numbers using repeated division method.
2. Write these numbers in the form of exponent and find the product of only those prime factors that have the highest power.
3. The product of these factors with the highest powers is the LCM of the given numbers

#### Example:

Find LCM of two positive integers 120 and 300.

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

## How to Find HCF and LCM of a Fraction?

HCF of a fraction:  HCF of Numerators /LCM of Denominators.

LCM of a fraction:  LCM of Numerators /HCF of Denominators.

### Example

Q. Find the HCF and LCM of 1/3 , 8/7, 9/11.

Solution: LCM of given numbers : LCM(1,8,9)/HCF (3,7,11) = 72/1.

HCF of given numbers : HCF(1,8,9)/LCM(3,7,11) = 1/231.

Sample Problems on HCF and LCM:

### Q1: 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.

### Q2: 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.

### Q3: 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.

### Q4: Find the largest number which 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.

### Q5: Find the least number which when divided by 5,7,9 and 12, leaves the same remainder 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

### Q6: 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

### Q7: 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.

### Q8: 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.

### Q9: 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.

### Q10: 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.

### Q11: Three rectangular fields having areas of 60 m2, 84 m2, 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.

### Q12: 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

### Q13: 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 24min.The bells will ring together again at 4 AM+ 24 minutes =4:24 AM.

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