(RRB JE CBT 1 Mathematics Chapter)
✳️ Concept Overview / अध्याय की झलक
English:
LCM (Least Common Multiple) and HCF (Highest Common Factor) are two important operations based on multiples and factors of numbers.
They form the foundation of many Railway JE Maths topics like Time, Work, Fractions, and Number System problems.
Hindi:
LCM (लघुत्तम समापवर्त्य) और HCF (महत्तम समापवर्तक) गणित के महत्वपूर्ण भाग हैं।
इनका प्रयोग अनेक प्रश्नों में होता है जैसे समय और कार्य, भिन्न, गति, और संख्या पद्धति से जुड़े प्रश्नों में।
🧮 Basic Definitions / मूल परिभाषाएँ
| Term | English Meaning | Hindi Meaning |
|---|---|---|
| LCM (Least Common Multiple) | The smallest number that is a multiple of each of the given numbers. | वह सबसे छोटी संख्या जो सभी दी गई संख्याओं से विभाज्य हो। |
| HCF (Highest Common Factor) | The largest number which divides each of the given numbers completely. | वह सबसे बड़ी संख्या जो सभी संख्याओं को पूर्णतः विभाजित करे। |
🔹 Prime Factorization Method / अभाज्य गुणनखंड विधि
Step-by-Step Method / चरणबद्ध विधि
English:
-
Break each number into its prime factors.
-
For LCM, take each prime factor with its highest power.
-
For HCF, take each common prime factor with its lowest power.
Hindi:
-
प्रत्येक संख्या को उसके अभाज्य गुणनखंडों में विभाजित करें।
-
LCM निकालने के लिए प्रत्येक गुणनखंड की सर्वाधिक घात (highest power) लें।
-
HCF निकालने के लिए सामान्य गुणनखंडों की न्यूनतम घात (lowest power) लें।
Example 1:
Find LCM and HCF of 24 and 36.
Prime factors:
24 = 2³ × 3¹
36 = 2² × 3²
LCM = 2³ × 3² = 72
HCF = 2² × 3¹ = 12
✅ Trick Tip:
✍️ “LCM × HCF = Product of the Numbers”
i.e., 24 × 36 = 864 and 72 × 12 = 864 ✅
⚡ Shortcut Method (LCM of Large Numbers) / शॉर्टकट विधि
When numbers are large and not easily factorized:
English:
Use division method – divide the numbers by small prime numbers (2, 3, 5, 7…) until all become 1. Multiply all divisors and remaining quotients = LCM.
Hindi:
यदि संख्याएँ बड़ी हों, तो विभाजन विधि अपनाएँ —
2, 3, 5, 7 आदि से बार-बार भाग दें जब तक सभी 1 न बन जाएँ।
सभी भाजकों और शेष संख्याओं का गुणनफल = LCM होगा।
Example 2:
Find LCM of 8, 12, and 18.
| Step | Numbers | Divisor |
|---|---|---|
| 1 | 8, 12, 18 | ÷ 2 → 4, 6, 9 |
| 2 | 4, 6, 9 | ÷ 2 → 2, 3, 9 |
| 3 | 2, 3, 9 | ÷ 3 → 2, 1, 3 |
| 4 | 2, 1, 3 | ÷ 3 → 2, 1, 1 |
| 5 | 2, 1, 1 | ÷ 2 → 1, 1, 1 |
✅ LCM = 2 × 2 × 3 × 3 = 36
🔢 Shortcut Tricks for Remainders / शेषफल निकालने के शॉर्टकट
Case 1: When same remainder is left
English:
If same remainder ‘r’ is left when dividing numbers by divisor ‘d’, then
Required Number = LCM of differences + remainder
Hindi:
यदि सभी संख्याओं को किसी भाजक से भाग देने पर समान शेषफल ‘r’ मिलता है, तो
आवश्यक संख्या = अंतरों का LCM + शेषफल
Example 3:
Find the smallest number which when divided by 8, 12, and 16 leaves remainder 5.
Differences = (8−5), (12−5), (16−5) = 3, 7, 11
LCM of 3, 7, 11 = 231
Answer = 231 + 5 = 236
Case 2: When numbers divide completely
English:
If the number divides exactly (remainder 0),
then the required number = LCM of divisors.
Hindi:
यदि सभी संख्याओं से भाग देने पर कोई शेषफल नहीं बचता,
तो आवश्यक संख्या = सभी भाजकों का LCM होगी।
Example 4:
Find the smallest number divisible by 6, 8, 9.
LCM = 72 ✅
💡 Fast Track Tricks / फास्ट ट्रैक ट्रिक्स
| Type | Trick | Example |
|---|---|---|
| LCM × HCF = Product of Numbers | Useful to find missing value | If LCM = 120, HCF = 6 → Product = 720 |
| When same remainder | Add remainder after LCM of differences | 8,12,16 leave remainder 5 → Ans = 231+5=236 |
| If numbers divide exactly | Required number = LCM | 6, 8, 9 → 72 |
| When two numbers are co-prime | LCM = Product | 7 & 9 → 63 |
| If one divides other | LCM = Greater, HCF = Smaller | 6,12 → LCM=12, HCF=6 |
🧠 Common Mistakes / सामान्य गलतियाँ
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❌ Forgetting to include highest power in LCM.
✅ Always take the highest exponent in each prime factor. -
❌ Confusing difference vs. remainder.
✅ For same remainder questions, take LCM of differences, not numbers. -
❌ Using LCM formula for HCF questions.
✅ Remember: LCM × HCF = Product → use only for two numbers.
🧾 Practice Questions / अभ्यास प्रश्न
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Find LCM and HCF of 18, 24, and 30.
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The LCM of two numbers is 180 and their HCF is 15. If one number is 45, find the other.
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Find the smallest number divisible by 8, 12, and 15.
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Find the least number which when divided by 9, 12, 15 leaves remainder 3.
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Find HCF of 84, 126, and 210 using prime factors.
✅ Answer Key / उत्तर कुंजी
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LCM = 360, HCF = 6
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(LCM × HCF) / One number = (180 × 15) / 45 = 60
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LCM = 120
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LCM of (9−3, 12−3, 15−3) = LCM(6,9,12)=36 → Ans = 36 + 3 = 39
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HCF = 42
🧭 Final Summary / सारांश
| Concept | Formula / Trick |
|---|---|
| LCM × HCF = Product | Use for 2 numbers |
| LCM (Prime factors) | Take highest powers |
| HCF (Prime factors) | Take lowest powers |
| Same Remainder Case | LCM of (difference) + remainder |
| Co-prime Numbers | LCM = Product, HCF = 1 |
🔖 Fast Track Note
🌀 “When numbers differ by a fixed difference, their LCM = Difference × (Quotient of highest multiple)”
👉 Useful for consecutive multiples like 8, 16, 24 → LCM = 8×3 = 24.
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