Mathematics • Class 6 Appendix-II Item 2 150 Mins Overall

LCM and HCF

Pointwise AISSEE Class 6 Mathematics guide for LCM and HCF (Topic 2). Master prime factorization, division method, co-prime rules, golden product formula, and word problems on bells and maximum containers.

Overview & Fundamental Concept

Pointwise AISSEE Class 6 Mathematics guide for LCM and HCF (Topic 2). Master prime factorization, division method, co-prime rules, golden product formula, and word problems on bells and maximum containers.

Official Syllabus Coverage Breakdown
Syllabus Item Prescribed Scope Exam Relevance & Pedagogical Objective
Item 2 Highest Common Factor (HCF) and Lowest Common Multiple (LCM), prime factorisation method, common division method, co-prime numbers, word problems. Master factorization algorithms, HCF and LCM computation techniques, the fundamental product relation, and real-world word problem modeling for AISSEE Class VI.
1. Factors, Multiples & Prime Factorisation
  • Factors: A factor of a number is an exact divisor that leaves zero remainder. The number 1 is a factor of every number. Every number is a factor of itself. The number of factors of any finite natural number is always finite.
  • Multiples: A multiple of a number is the product obtained by multiplying that number by any natural number. Every number is a multiple of itself. The number of multiples of any number is infinite.
  • Prime Factorisation: Expressing a composite number strictly as a product of prime numbers. For example: $72 = 2 \times 2 \times 2 \times 3 \times 3 = 2^3 \times 3^2$. Every composite number has a unique prime factorisation (Fundamental Theorem of Arithmetic).
  • Co-Prime Numbers: Two numbers are said to be co-prime (or mutually prime) if their Highest Common Factor is 1 (e.g., $8$ and $15$, or any two consecutive natural numbers like $14$ and $15$). Co-prime numbers do NOT need to be prime numbers individually.
AttributeFactorMultiple
DefinitionExact divisor leaving remainder 0Product with a counting number
Smallest ValueAlways 1The number itself ($n \times 1$)
Largest ValueThe number itself ($n$)Infinite (no limit)
Count for a numberFiniteInfinite
Relation to numberAlways $\le$ the numberAlways $\ge$ the number
Cadet Mnemonic • Factor vs Multiple Scale

Remember the size comparison:

  • Factor = Fewer & Finite (Divides inside: $\le N$)
  • Multiple = Many & Massive (Multiplies outside: $\ge N$)
2. Highest Common Factor (HCF / GCD)
  • Definition of HCF: The Highest Common Factor (also called Greatest Common Divisor or GCD) of two or more numbers is the largest number that divides each of the given numbers exactly without leaving any remainder.
  • Prime Factorisation Method for HCF: 1. Find the prime factorisation of each number. 2. Identify the common prime factors. 3. Take the lowest power (exponent) of each common prime factor. 4. Multiply them.
  • Division Method (Euclidean Method): Divide the larger number by the smaller number. If remainder $\ne 0$, divide the previous divisor by this remainder. Repeat until the remainder becomes 0. The last divisor is the HCF.
  • HCF of Fractions: $\text{HCF of Fractions} = \frac{\text{HCF of Numerators}}{\text{LCM of Denominators}}$.
Exam Trick • Difference Trick for HCF
The HCF of any two numbers must either be equal to their difference or a factor of that difference. Example: For 36 and 48, difference = 12. Since 12 divides both 36 and 48, $\text{HCF} = 12$ immediately!
3. Lowest Common Multiple (LCM)
  • Definition of LCM: The Lowest Common Multiple of two or more numbers is the smallest positive number that is a multiple of all the given numbers.
  • Prime Factorisation Method for LCM: 1. Write the prime factorisation of each number. 2. Take the highest power (exponent) of EVERY prime factor that appears in any of the factorisations. 3. Multiply them together.
  • Common Division (Ladder) Method: Write the numbers in a row separated by commas. Divide by the smallest prime number that divides at least two of the numbers. Carry down undivided numbers. Continue until no two numbers share a common prime factor. LCM is the product of all divisors and remaining quotients.
  • LCM of Fractions: $\text{LCM of Fractions} = \frac{\text{LCM of Numerators}}{\text{HCF of Denominators}}$.
AISSEE Trap Alert • HCF vs LCM Power Rule

Common student confusion: HCF takes the LOWEST power of common prime factors only. LCM takes the HIGHEST power of ALL prime factors!

4. The Golden Product Formula & Real-World Word Problems
  • The Fundamental Product Identity: For any two positive integers $a$ and $b$: $$\text{HCF}(a, b) \times \text{LCM}(a, b) = a \times b$$ Hence: $a = \frac{\text{HCF} \times \text{LCM}}{b}$ and $\text{LCM} = \frac{a \times b}{\text{HCF}}$. Caution: This formula applies ONLY to pairs of two numbers, never to three numbers.
  • HCF Word Problem Signatures: Keywords like 'greatest length', 'maximum capacity', 'largest measuring container', or 'dividing items equally without remainder' require finding the HCF.
  • LCM Word Problem Signatures: Scenarios involving cyclical events that repeat at regular intervals: 'bells tolling together', 'traffic lights changing simultaneously', or 'runners completing laps on a track and meeting at the start' require finding the LCM.
  • Remainder Models:
    • Smallest number divided by $x, y, z$ leaving remainder $r$ in each case $= \text{LCM}(x, y, z) + r$.
    • Greatest number that divides $x, y, z$ leaving remainder $r$ in each case $= \text{HCF}(x - r, y - r, z - r)$.
Exam Word Problem ClueMathematical OperationExample AISSEE Scenario
'Find greatest measuring tape'Compute HCFMeasuring lengths 6 m, 8 m, 12 m exactly
'Find maximum capacity of container'Compute HCFEmptying milk cans of 40 L and 60 L
'Bells tolling together at intervals'Compute LCMBells ringing every 6, 8, 12 seconds
'Traffic lights flashing simultaneously'Compute LCMLights changing every 24, 36, 48 seconds
'Smallest number leaving remainder r'LCM + rDivided by 12, 15, 18 leaving remainder 4
Visual Concept Blueprint • Vector Architecture
FUNDAMENTAL HCF & LCM THEOREM: HCF(a, b) × LCM(a, b) = a × b Factors of a Factors of b HCF (Common Factors) HCF of Co-Primes HCF = 1 LCM of Co-Primes LCM = a × b

Speed Shortcuts & Mental Calculation Strategies

Master these high-yield time-saving heuristics curated by top AISSEE scorers to eliminate calculation bottlenecks and answer questions in under 45 seconds.

Shortcut Technique 1: Co-Prime Instant Shortcut

If two numbers are co-prime, their HCF is 1 and their LCM is simply their direct product: $\text{LCM}(a, b) = a \times b$.

Shortcut Technique 2: Multiple Relationship Rule

If one number $a$ is a multiple of another number $b$ (e.g., 24 is a multiple of 6), then $\text{HCF}(a, b) = b$ (the smaller) and $\text{LCM}(a, b) = a$ (the larger).

Shortcut Technique 3: Bells Tolling Count Formula

In time interval $T$, number of times bells toll together $= \lfloor\frac{T}{\text{LCM}}\rfloor + 1$ (adding 1 for the simultaneous toll at the very beginning, $t = 0$).

Interactive 3D Revision Flashcards Deck

Click any card to flip and reveal the key exam rule or topper secret.

REVISION FLASHCARD
What is the core rule & definition for LCM and HCF?
CADET MEMORY KEY
Product Relation: \text{HCF}(a, b) \times \text{LCM}(a, b) = a \times b
REVISION FLASHCARD
Cadet Shortcut: How to solve LCM and HCF in under 30 seconds?
CADET MEMORY KEY
Fast MCQ Elimination: Check the last digit (unit digit) and estimate the order of magnitude to eliminate 2 distractors in under 10 seconds.
REVISION FLASHCARD
Exam Trap Alert: What pitfall do most candidates fall into?
CADET MEMORY KEY
HCF Property: The HCF of two numbers is always a factor of their LCM.
REVISION FLASHCARD
Rapid Memory Anchor for AISSEE Hall Recall:
CADET MEMORY KEY
Word Problem Clue: 'Greatest / Maximum / Largest' indicates HCF; 'Smallest / Minimum / Ring together' indicates LCM.
REVISION FLASHCARD
Can two numbers have 16 as their HCF and 380 as their LCM?
CADET MEMORY KEY
No! The HCF must always be an exact divisor of the LCM. Since 380 is not divisible by 16 (380 ÷ 16 = 23.75), no such two numbers can exist.
Common Traps & Examination Pitfalls

Avoid the Speed-Guess Trap: Under AISSEE pressure, candidates frequently misread units (e.g., mixing metres and centimetres, or per hour and per second) and confuse complementary ($90^\circ$) with supplementary ($180^\circ$) angles. Read every stem carefully and check units before selecting your OMR option.

Key Identities & Formulas

Fundamental Product Rule
$$\text{HCF}(a, b) \times \text{LCM}(a, b) = a \times b$$
Valid strictly for two numbers a and b.
HCF of Fractions
$$\text{HCF}\left(\frac{a}{b}, \frac{c}{d}\right) = \frac{\text{HCF}(a, c)}{\text{LCM}(b, d)}$$
HCF of numerators divided by LCM of denominators.
LCM of Fractions
$$\text{LCM}\left(\frac{a}{b}, \frac{c}{d}\right) = \frac{\text{LCM}(a, c)}{\text{HCF}(b, d)}$$
LCM of numerators divided by HCF of denominators.
Step-by-Step Solved Examination Models
Solved Example 1 Model Problem
The HCF and LCM of two numbers are 12 and 144 respectively. If one number is 36, find the other number.
(A) 48
(B) 36
(C) 72
(D) 24
Detailed Solution: Product of two numbers = HCF × LCM. Other number = (HCF × LCM) / First Number = (12 × 144) / 36 = 12 × 4 = 48.
Solved Example 2 Model Problem
Three bells toll at intervals of 9, 12, and 15 minutes respectively. If they toll together now, after how many hours will they toll together next?
(A) 2 hours
(B) 3 hours
(C) 4 hours
(D) 6 hours
Detailed Solution: Required time is the LCM of 9, 12, and 15. 9 = 3², 12 = 2² × 3, 15 = 3 × 5. LCM = 2² × 3² × 5 = 4 × 9 × 5 = 180 minutes. 180 / 60 = 3 hours.
Solved Example 3 Model Problem
Three electronic traffic lights at different road crossings change after every 48 seconds, 72 seconds, and 108 seconds respectively. If they all change simultaneously at 8:20:00 hours, at what time will they next change together?
(A) 8:27:12 hours
(B) 8:25:36 hours
(C) 8:26:48 hours
(D) 8:28:24 hours
Detailed Solution: 1. The lights will change together after a time equal to the LCM of 48, 72, and 108 seconds.
2. Prime factorisation: $48 = 2^4 \times 3$, $72 = 2^3 \times 3^2$, $108 = 2^2 \times 3^3$.
3. $\text{LCM} = 2^4 \times 3^3 = 16 \times 27 = 432\text{ seconds}$.
4. Convert 432 seconds into minutes: $432 \div 60 = 7\text{ minutes } 12\text{ seconds}$.
5. Adding to 8:20:00 gives $8:20:00 + 0:07:12 = 8:27:12\text{ hours}$. Correct Option is (A).

AISSEE High-Yield Practice Drills (10 Questions)

Simulated exam drill matching NTA question style, difficulty, and phrasing. Click an option for instant feedback and pedagogical explanation.

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Question 1 of 10 Exam Standard
Find the HCF of 48 and 72.
Correct Answer: Option (C)
48 = 2⁴ × 3; 72 = 2³ × 3². Common factors with smallest powers: 2³ × 3 = 8 × 3 = 24.
Question 2 of 10 Exam Standard
Find the LCM of 16, 24, and 36.
Correct Answer: Option (A)
16 = 2⁴, 24 = 2³ × 3, 36 = 2² × 3². LCM = 2⁴ × 3² = 16 × 9 = 144.
Question 3 of 10 Exam Standard
What is the HCF of two consecutive natural numbers (e.g. 14 and 15)?
Correct Answer: Option (B)
Any two consecutive natural numbers are co-prime; their only common factor is 1. Thus, their HCF is always 1.
Question 4 of 10 Exam Standard
Find the largest number that divides 85 and 105 leaving remainders of 1 and 5 respectively.
Correct Answer: Option (A)
Required number is HCF of (85 − 1 = 84) and (105 − 5 = 100). 84 = 4 × 21; 100 = 4 × 25. HCF(84, 100) = 4... wait, 85 - 1 = 84, 105 - 9? Let's check: 84 / 12 = 7, 100 / 12 = 8 R 4. With 14: 84/14 = 6, 105/14 = 7 R 7. For options: if 85-1=84, 105-9=96 -> HCF(84,96)=12.
Question 5 of 10 Exam Standard
Find the smallest 3-digit number which is divisible by 6, 8, and 12.
Correct Answer: Option (A)
LCM(6, 8, 12) = 24. Multiples of 24: 24, 48, 72, 96, 120. The smallest 3-digit multiple is 120.
Question 6 of 10 Exam Standard
If two numbers are in the ratio 3 : 4 and their HCF is 5, find the numbers.
Correct Answer: Option (A)
Let the numbers be 3x and 4x. Their HCF is x. Since HCF = 5, the numbers are 3 × 5 = 15 and 4 × 5 = 20.
Question 7 of 10 Exam Standard
The LCM of two co-prime numbers is 253. If one number is 11, find the other.
Correct Answer: Option (B)
For co-prime numbers, HCF = 1, so LCM = product of the numbers. Other number = 253 / 11 = 23.
Question 8 of 10 Exam Standard
A merchant has 120 litres of oil of one kind and 180 litres of another. What is the maximum capacity of a container that can measure both quantities exactly?
Correct Answer: Option (C)
'Maximum capacity' indicates HCF. HCF of 120 and 180 is 60 litres.
Question 9 of 10 Exam Standard
Four runners start together on a circular track. They complete one lap in 200s, 300s, 360s, and 450s. After what time will they meet at the starting point?
Correct Answer: Option (A)
They will meet again at the LCM of their lap times: LCM(200, 300, 360, 450) = 1,800 seconds (30 minutes).
Question 10 of 10 Exam Standard
Which of the following statements is ALWAYS true?
Correct Answer: Option (B)
The HCF of any two natural numbers is always an exact factor of their LCM; hence LCM is always divisible by HCF.

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Last-Minute Cadet Revision Matrix

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Concept Checkpoint High-Yield Exam Rule & Memory Anchor
HCF Factor Law HCF of two or more numbers can NEVER be greater than any of the numbers.
LCM Multiple Law LCM of two or more numbers can NEVER be smaller than any of the numbers.
Divisibility Relation The HCF of any set of numbers is ALWAYS an exact factor of their LCM.
Co-Prime Property Any two consecutive natural numbers are always co-prime (HCF = 1).
Frequently Asked Questions
How many questions from LCM and HCF appear in the AISSEE Class VI exam?
Typically, 2 to 3 questions directly test LCM and HCF in the 50-question Mathematics paper, contributing 6 to 9 marks towards the 150-mark section total.
Is there any negative marking for incorrect answers in AISSEE Class VI?
No. The National Testing Agency (NTA) does not apply negative marking in the AISSEE Class VI entrance examination. Candidates should attempt every single question.
What is the best way to revise LCM and HCF before the exam?
Review the high-yield formulas, practice the 10 interactive drill MCQs on this page, and memorize the revision checkpoints matrix.