In number theory, two important concepts that help us understand the relationship between numbers are the Highest Common Factor (HCF) and the Least Common Multiple (LCM). These concepts are not just theoretical; they have practical applications in daily life and competitive exams like the Delhi Police Constable (Executive) Examination.
HCF is the largest number that divides two or more numbers exactly, without leaving a remainder. It helps in simplifying fractions and solving problems related to divisibility.
LCM is the smallest number that is a multiple of two or more numbers. It is useful in scheduling events, adding fractions with different denominators, and solving problems involving repeated cycles.
Understanding how to find HCF and LCM efficiently can save time during exams and help solve a variety of numerical problems. We will explore different methods to find HCF and LCM, including prime factorization, the division method, and the Euclidean algorithm.
Prime factorization is the process of expressing a number as a product of its prime factors. A prime number is a number greater than 1 that has no divisors other than 1 and itself. For example, 2, 3, 5, 7, 11 are prime numbers.
Breaking a number into prime factors helps us find the HCF and LCM by comparing the prime factors of the numbers involved.
Let's see how to find the prime factors of 60 using a factor tree:
Explanation:
Thus, the prime factorization of 60 is:
Or written with exponents:
The division method is a systematic way to find both HCF and LCM by dividing numbers simultaneously by their common prime factors until no further division is possible.
Let's illustrate this with the numbers 48 and 60:
Stepwise explanation:
The HCF is the product of all the divisors used: \(2 \times 2 \times 3 = 12\).
The LCM is the product of all the divisors and the remaining numbers: \(2 \times 2 \times 3 \times 4 \times 5 = 240\).
The Euclidean algorithm is an efficient method to find the HCF of two numbers using repeated division and remainders. It is especially useful for large numbers where prime factorization is difficult.
The algorithm is based on the principle that the HCF of two numbers also divides their difference.
Here is a flowchart illustrating the Euclidean algorithm to find the HCF of 252 and 105:
graph TD A[Start with numbers 252 and 105] B[Divide 252 by 105] C[Find remainder r = 252 mod 105 = 42] D[Replace 252 with 105, 105 with r (42)] E{Is remainder 0?} F[Yes: HCF is 42] G[No: Repeat division with new pair] A --> B --> C --> D --> E E -- No --> G --> B E -- Yes --> FStepwise explanation:
Step 1: Find prime factors of 36.
36 = 2 x 2 x 3 x 3 = \(2^2 \times 3^2\)
Step 2: Find prime factors of 48.
48 = 2 x 2 x 2 x 2 x 3 = \(2^4 \times 3^1\)
Step 3: Find common prime factors with lowest powers for HCF.
Common prime factors: 2 and 3
Lowest powers: \(2^2\) and \(3^1\)
HCF = \(2^2 \times 3^1 = 4 \times 3 = 12\)
Step 4: Find all prime factors with highest powers for LCM.
Highest powers: \(2^4\) and \(3^2\)
LCM = \(2^4 \times 3^2 = 16 \times 9 = 144\)
Answer: HCF = 12, LCM = 144
Step 1: Divide 252 by 105.
252 / 105 = 2 remainder 42
Step 2: Now find HCF of 105 and 42.
105 / 42 = 2 remainder 21
Step 3: Now find HCF of 42 and 21.
42 / 21 = 2 remainder 0
Step 4: Since remainder is 0, HCF is 21.
Answer: HCF(252, 105) = 21
Step 1: Recall the formula:
Step 2: Substitute the known values:
5 x LCM = 15 x 20 = 300
Step 3: Solve for LCM:
LCM = \(\frac{300}{5} = 60\)
Answer: LCM of 15 and 20 is 60
Step 1: Write the numbers side by side:
48, 60, 72
Step 2: Divide by 2 (common prime factor):
48 / 2 = 24, 60 / 2 = 30, 72 / 2 = 36
Step 3: Divide again by 2:
24 / 2 = 12, 30 / 2 = 15, 36 / 2 = 18
Step 4: Divide by 3:
12 / 3 = 4, 15 / 3 = 5, 18 / 3 = 6
Step 5: Divide by 2:
4 / 2 = 2, 5 / 2 = not divisible, stop dividing by 2
Step 6: Divide by 3:
2 / 3 = no, 5 / 3 = no, 6 / 3 = 2 (only one divisible, so stop)
Step 7: Remaining numbers are 2, 5, and 6
HCF: Multiply all common divisors: \(2 \times 2 \times 3 = 12\)
LCM: Multiply all divisors and remaining numbers:
LCM = \(2 \times 2 \times 3 \times 2 \times 5 \times 6 = 720\)
Answer: HCF = 12, LCM = 720
Step 1: Identify the problem as finding the LCM of 12 and 18.
Step 2: Find prime factorization:
12 = \(2^2 \times 3\)
18 = \(2 \times 3^2\)
Step 3: Take highest powers of each prime for LCM:
LCM = \(2^2 \times 3^2 = 4 \times 9 = 36\)
Answer: The events will occur together again after 36 days.
When to use: When prime factorization is time-consuming or difficult due to large numbers.
When to use: When either HCF or LCM is given and you need to find the other quickly.
When to use: When learning prime factorization or working with smaller numbers.
When to use: When dealing with three or more numbers to avoid complicated factorization.
When to use: To speed up calculations and avoid unnecessary steps.
Progress tracking is paywalled — subscribe to mark subtopics as understood and save your streak.
Go to practice →