LCM & HCF: Factorization, Division & Applicationsल०स० एवं म०स०: गुणनखण्ड, भाग विधि एवं अनुप्रयोग
Mastery of Lowest Common Multiple (LCM) and Highest Common Factor (HCF) through prime factorization, division algorithms, fraction analysis, and real-world periodic event applications. High examination weightage in UP Super TET Junior Part 2C for testing analytical problem-solving speed under -1 negative marking constraints.
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Syllabus Focus & High-Yield Strategy: Mastery of Lowest Common Multiple (LCM) and Highest Common Factor (HCF) through prime factorization, division algorithms, fraction analysis, and real-world periodic event applications. High examination weightage in UP Super TET Junior Part 2C for testing analytical problem-solving speed under -1 negative marking constraints.
1. Fundamental Axioms, Definitions & Foundational Properties
HPL — Highest uses Powers Least (for HCF, take lowest exponent of common primes); Lowest uses Powers Largest (for LCM, take highest exponent of all primes).2. Formulas, Standard Theorems & Algorithmic Procedures
| Category / Case | Formula / Mathematical Rule | Topper Calculation Shortcut |
|---|---|---|
| Two Numbers Product | $ ext{HCF}(a,b) imes ext{LCM}(a,b) = a imes b$ | Find HCF first via division, then divide product $(a imes b)$ by HCF to get LCM instantly. |
| Fractions HCF & LCM | $ ext{HCF}left( rac{a}{b}, rac{c}{d} ight) = rac{ ext{HCF}(a, c)}{ ext{LCM}(b, d)}$ | Numerator operation matches the outer function name (HCF numerator, LCM denominator). |
| Fractions LCM & HCF | $ ext{LCM}left( rac{a}{b}, rac{c}{d} ight) = rac{ ext{LCM}(a, c)}{ ext{HCF}(b, d)}$ | Denominator operation is always the opposite of the outer function. |
| Decimal Numbers | Convert all decimals to like fractions by appending zeros, compute HCF/LCM, then adjust decimal point. | Count maximum decimal places $k$, multiply by $10^k$, compute, and divide final result by $10^k$. |
3. Word Problem Modeling & Practical Applications
4. Mathematical Edge Cases, Sign Conventions & Common Traps
SALI — Subtract After LCM is found (for remainder problems where remainder is added at the end); Subtract Before HCF (for remainder problems where remainder is subtracted first).Comprehensive Mathematical Matrix & Method Comparison
| Parameter / Dimension | Method / Concept A | Method / Concept B | Exam Speed & Accuracy Takeaway |
|---|---|---|---|
| Prime Factorization vs Division | Prime Factorization: Best for small numbers (< 1000) and algebraic expressions. | Division Method (Euclidean): Best for large numbers and successive remainders. | Use Prime Factorization for algebraic/variable terms; use Division for large numerical values. |
| HCF Calculation Properties | HCF of two numbers divides their sum, difference, and LCM. | HCF is never greater than the smaller of the two numbers. | Use difference method: $ ext{HCF}(a, b) = ext{HCF}(a, b-a)$ to quickly reduce large numbers. |
| LCM Calculation Properties | LCM of two coprime numbers is their direct product. | LCM is always a multiple of the HCF of those numbers. | Always check if smaller number divides larger number; if yes, larger number is LCM. |
| Application Contexts | HCF: Partitioning, measuring, maximum capacity, highest common divisor. | LCM: Synchronization, periodic recurrence, simultaneous bells, circular tracks. | Match keywords in question stem ('greatest' $ o$ HCF, 'least/together' $ o$ LCM) instantly. |
1. What is the HCF of two numbers whose LCM is $2310$ and product is $48510$?
Using the standard property relating HCF, LCM, and product of two numbers: $ ext{HCF}(a, b) imes ext{LCM}(a, b) = a imes b$. Substituting the given values: $ ext{HCF} imes 2310 = 48510$. Therefore, $ ext{HCF} = rac{48510}{2310} = rac{4851}{231} = 21$. Distractors like $11$ or $42$ arise from incorrect division or arithmetic calculation errors.
2. Find the greatest number that divides 414, 751, and 1014, leaving remainders 7, 11, and 15, respectively.
Subtract each remainder: 414-7=407=11 x 37, 751-11=740=20 x 37, and 1014-15=999=27 x 37. Since gcd(11,20,27)=1, the greatest common divisor is 37. The answer is Option B.
3. The LCM of two numbers is $12$ times their HCF. The sum of LCM and HCF is $403$. If one number is $93$, what is the other number?
Let HCF be $H$ and LCM be $L$. Given $L = 12H$ and $L + H = 403$. Substituting $L$: $12H + H = 403 implies 13H = 403 implies H = rac{403}{13} = 31$. Thus, $L = 12 imes 31 = 372$. Using the product formula: $a imes b = H imes L implies 93 imes b = 31 imes 372$. Solving for $b$: $b = rac{31 imes 372}{93} = rac{372}{3} = 124$. Distractors like $134$ or $144$ result from arithmetic division errors.
4. Three alarm clocks ring at intervals of $10$, $15$, and $20$ minutes respectively. If they ring together at $10:00 ext{ AM}$, at what time will they next ring together?
To find when the clocks ring together again, we calculate the LCM of the intervals: $ ext{LCM}(10, 15, 20)$. Prime factorization: $10 = 2 imes 5$, $15 = 3 imes 5$, $20 = 2^2 imes 5$. Taking highest powers of all primes: $ ext{LCM} = 2^2 imes 3 imes 5 = 4 imes 3 imes 5 = 60 ext{ minutes}$ (or $1 ext{ hour}$). Adding $1 ext{ hour}$ to $10:00 ext{ AM}$ gives $11:00 ext{ AM}$. Distractors like $10:30 ext{ AM}$ or $11:30 ext{ AM}$ arise from incorrect LCM calculation.
5. What is the LCM of $\frac{2}{3}, \frac{4}{9},$ and $\frac{5}{6}$?
The formula for the LCM of fractions is $ ext{LCM}left( rac{a}{b}, rac{c}{d}, rac{e}{f} ight) = rac{ ext{LCM}(a, c, e)}{ ext{HCF}(b, d, f)}$. Numerators are $2, 4, 5$; $ ext{LCM}(2, 4, 5) = 20$. Denominators are $3, 9, 6$; $ ext{HCF}(3, 9, 6) = 3$. Therefore, $ ext{LCM} = rac{20}{3}$. Distractors like $ rac{10}{3}$ or $ rac{20}{9}$ result from swapping numerator and denominator operations.
6. Find the least number which when divided by $12, 16,$ and $24$ leaves a remainder $7$ in each case.
First, find the LCM of the divisors $12, 16,$ and $24$. Prime factorization: $12 = 2^2 imes 3$, $16 = 2^4$, $24 = 2^3 imes 3$. $ ext{LCM}(12, 16, 24) = 2^4 imes 3 = 16 imes 3 = 48$. Since the number leaves a constant remainder $7$ in each case, we add the remainder to the LCM: $48 + 7 = 55$. Testing $55$: $55 div 12 = 4$ remainder $7$; $55 div 16 = 3$ remainder $7$; $55 div 24 = 2$ remainder $7$. Distractor $47$ is obtained by subtracting instead of adding.
7. The ratio of two numbers is $3:4$ and their HCF is $5$. What is their LCM?
Let the two numbers be $3x$ and $4x$. Since their ratio is in simplest form, their HCF is $x$. Given $ ext{HCF} = 5$, the numbers are $3 imes 5 = 15$ and $4 imes 5 = 20$. The LCM of two numbers is given by $ ext{HCF} imes ext{ratio}_1 imes ext{ratio}_2 = 5 imes 3 imes 4 = 60$. Alternatively, $ ext{LCM} = rac{15 imes 20}{ ext{HCF}(15, 20)} = rac{300}{5} = 60$. Distractor $45$ or $75$ arises from incorrect multiplication.
8. What is the HCF of $0.63, 1.05,$ and $2.1$?
Make all numbers have the same number of decimal places: $0.63, 1.05, 2.10$. Consider the integers formed by removing the decimal points: $63, 105, 210$. Find the HCF of $63, 105, 210$. Prime factorization: $63 = 3^2 imes 7$, $105 = 3 imes 5 imes 7$, $210 = 2 imes 3 imes 5 imes 7$. Common prime factors with lowest powers are $3^1 imes 7^1 = 21$. Since numbers had two decimal places, divide by $100$: $ rac{21}{100} = 0.21$. Distractor $0.07$ is obtained by taking the wrong common factor ($7$ only).
9. Find the greatest four-digit number which is exactly divisible by $15, 21,$ and $36$.
The LCM of 15, 21, and 36 is 2^2 x 3^2 x 5 x 7 = 1260. The greatest four-digit multiple is 1260 x 7 = 8820; the next multiple is five digits. The answer is Option A.
10. The sum of two numbers is $216$ and their HCF is $27$. How many such pairs of numbers can be formed?
Let the two numbers be $27a$ and $27b$, where $a$ and $b$ are coprime integers ($ ext{HCF}(a, b) = 1$). Given their sum is $216$: $27a + 27b = 216 implies 27(a + b) = 216 implies a + b = rac{216}{27} = 8$. We need to find pairs of coprime integers whose sum is $8$. Possible pairs for $(a, b)$: $(1, 7)$ and $(3, 5)$. (Note: $(2, 6)$ and $(4, 4)$ are not coprime since their HCF is greater than $1$). Thus, exactly $2$ such pairs can be formed: $(27 imes 1, 27 imes 7) = (27, 189)$ and $(27 imes 3, 27 imes 5) = (81, 135)$. Distractors like $3$ or $4$ arise from including non-coprime pairs.
5-Minute Exam Simulation Mode
Experience the real pressure of Super TET Junior (Part 2C Mathematics). You will face 10 randomized speed-math questions with negative marking (-1 for incorrect answers) and a strict 5-minute countdown.
High-Yield Rapid Recall Formulas & Laws (Must-Remember Pointers)
Speed Math & Memory Mnemonics Vault (Exam Retention Anchors)
HCF & LCM Power Selection Rule
HPL-LPL
H - Highest Common factor uses Powers Least; L - Lowest Common Multiple uses Powers Largest.
Remainder Word Problem Strategy
LARS
L - LCM Add remainder at end; S - Subtract remainder before finding HCF.
Negative Marking Traps & Common Blunders (-1 Penalty Avoidance)
Negative Marking Trap 1: Confusing HCF and LCM formulas for fractions (applying LCM to numerator when question asks for HCF).
Negative Marking Trap 2: Failing to check if generated pairs of numbers are coprime when solving sum/HCF ratio problems.
Negative Marking Trap 3: Adding the remainder *before* calculating LCM instead of adding it *after* the LCM calculation.
Negative Marking Trap 4: Forgetting to adjust decimal place values when calculating HCF or LCM of decimal numbers.