LCM Using Prime Factorisation for CBSE Class 10 Mathematics

The Lowest Common Multiple (LCM) of two or more positive integers is the smallest positive integer that is divisible by all of them without a remainder. Understanding how to calculate the LCM using the prime factorisation method is a core requirement of CBSE Class 10 Mathematics.

Beyond purely algebraic computations, LCM plays a vital role in solving real-world word problems involving periodic events, such as bells ringing simultaneously, runners completing circular laps, or traffic lights changing at synchronized intervals.

What You Will Learn

• Definition and conceptual foundation of LCM
• The greatest powers rule for prime factorisation
• Step-by-step method to compute LCM of two and three numbers
• Solving algebraic variable LCM problems
• Solving real-life word problems (circular tracks, traffic signals, alarm clocks)
• Verification using the relationship $\text{HCF} \times \text{LCM} = a \times b$
• Common student mistakes to avoid

1. What is LCM?

The Lowest Common Multiple (LCM) of two or more positive integers is the smallest positive integer that is a multiple of every one of the numbers.

Prime Factorisation Rule for LCM

LCM is the product of the greatest power of each prime factor involved in the numbers.

$\text{LCM}(a, b) = \prod_{i} p_i^{\max(a_i, b_i)}$

Important: <u>Unlike HCF, which takes only common prime factors, LCM includes EVERY prime factor that appears in ANY of the numbers, raised to its highest observed power.</u>

2. Step-by-Step Procedure

To find the LCM of numbers using prime factorisation:

1. Step 1: Write the prime factorisation of each number in exponential form.
2. Step 2: List all unique prime factors that appear in any of the factorisations.
3. Step 3: For each prime factor, identify its highest exponent across all numbers.
4. Step 4: Multiply these highest powers together to obtain the LCM.

3. Solved Numerical Examples

Solved Example 1: Two Numbers

Problem: Find the LCM of $96$ and $404$ using prime factorisation. Hence, verify that $\text{HCF} \times \text{LCM} = \text{Product of the two numbers}$.

Solution:

• Step 1 (Prime factorisation): $96 = 2^5 \times 3^1$ $404 = 2^2 \times 101^1$
• Step 2 (Identify all unique primes): The primes involved are $2$, $3$, and $101$.
• Step 3 (Select greatest powers):
  • Highest power of $2$: $\max(5, 2) = 5 \implies 2^5 = 32$
  • Highest power of $3$: $\max(1, 0) = 1 \implies 3^1 = 3$
  • Highest power of $101$: $\max(0, 1) = 1 \implies 101^1 = 101$
• Step 4 (Compute LCM): $\text{LCM}(96, 404) = 2^5 \times 3 \times 101 = 32 \times 3 \times 101 = 9696$

Verification:

• Calculate HCF (smallest powers of common primes): $\text{HCF}(96, 404) = 2^{\min(5, 2)} = 2^2 = 4$
• Check product: $\text{HCF} \times \text{LCM} = 4 \times 9696 = 38784$ $\text{Product of numbers} = 96 \times 404 = 38784$
• Since $\text{HCF} \times \text{LCM} = a \times b$, the relationship is verified.

Solved Example 2: Algebraic Variables (CBSE Board Question)

Problem: If two positive integers $p$ and $q$ can be expressed as $p = a b^2$ and $q = a^3 b$, where $a$ and $b$ are prime numbers, find $\text{LCM}(p, q)$.

Solution:

• We are given: $p = a^1 \times b^2$ $q = a^3 \times b^1$
• Primes involved: $a$ and $b$.
• Highest power of each factor:
  • For prime $a$: $\max(1, 3) = 3 \implies a^3$
  • For prime $b$: $\max(2, 1) = 2 \implies b^2$
• Therefore: $\text{LCM}(p, q) = a^3 b^2$

4. Real-World Word Problems (CBSE Board Classics)

Word Problem: Circular Sports Track

Problem: There is a circular path around a sports field. Sonia takes $18$ minutes to drive one round of the field, while Ravi takes $12$ minutes for the same. Suppose they both start at the same point and at the same time, and go in the same direction. After how many minutes will they meet again at the starting point?

Solution:

1. Sonia returns to the start line at multiples of $18$ minutes: $18, 36, 54, 72, \dots$
2. Ravi returns to the start line at multiples of $12$ minutes: $12, 24, 36, 48, \dots$
3. They will meet at the starting point again after a duration that is a common multiple of $18$ and $12$. To find the first time they meet, we need the Least Common Multiple (LCM) of $18$ and $12$.
4. Prime factorise both times: $18 = 2 \times 3^2$ $12 = 2^2 \times 3$
5. Compute LCM: $\text{LCM}(18, 12) = 2^{\max(1, 2)} \times 3^{\max(2, 1)} = 2^2 \times 3^2 = 4 \times 9 = 36$
6. Therefore, <u>Sonia and Ravi will meet again at the starting point after $36$ minutes</u>.

5. HCF vs LCM Comparison

Remember: For any two numbers $a$ and $b$, their $\text{HCF}(a, b)$ is always a factor of their $\text{LCM}(a, b)$. If HCF does not divide LCM evenly, an arithmetic error has occurred!

Common Mistake: Confusing HCF with LCM in word problems. If the question asks for when recurring events will coincide next, you must always find LCM!