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HCF, LCM, and Number Properties - Aptitude Mastery Guide

Section titled “HCF, LCM, and Number Properties - Aptitude Mastery Guide”

Category: Quantitative Aptitude
Generated on: 2025-07-15 09:15:16
Source: Aptitude Mastery Guide Generator

HCF, LCM, and Number Properties: A Comprehensive Guide

Section titled “HCF, LCM, and Number Properties: A Comprehensive Guide”

This guide is designed to be your comprehensive reference for HCF, LCM, and Number Properties, essential topics in quantitative aptitude. We’ll cover foundational concepts, key tricks, essential formulas, solved examples, and practice problems to help you master this area and excel in your exams.

1. Foundational Concepts

Section titled “1. Foundational Concepts”

a) Factors and Multiples:

  • A factor of a number divides the number exactly without leaving a remainder. For example, the factors of 12 are 1, 2, 3, 4, 6, and 12.
  • A multiple of a number is obtained by multiplying the number by an integer. For example, the multiples of 3 are 3, 6, 9, 12, and so on.

b) Prime and Composite Numbers:

  • A prime number is a natural number greater than 1 that has exactly two distinct positive divisors: 1 and itself. Examples: 2, 3, 5, 7, 11, 13, etc. The number 2 is the only even prime number.
  • A composite number is a natural number greater than 1 that has more than two distinct positive divisors. Examples: 4, 6, 8, 9, 10, etc.
  • The number 1 is neither prime nor composite.

c) Highest Common Factor (HCF) / Greatest Common Divisor (GCD):

  • The HCF of two or more numbers is the largest number that divides each of them exactly. It’s also known as the Greatest Common Divisor (GCD).
  • Why it matters: HCF is crucial for simplifying fractions, dividing quantities equally, and solving problems related to remainders.
  • Understanding the ‘why’: The HCF represents the largest common ‘building block’ present in all the numbers.

d) Least Common Multiple (LCM):

  • The LCM of two or more numbers is the smallest number that is exactly divisible by each of them.
  • Why it matters: LCM is essential for solving problems involving time and work, ratios, and finding common points in cyclical events.
  • Understanding the ‘why’: The LCM represents the smallest number that can be constructed using multiples of all the given numbers.

e) Co-prime Numbers:

  • Two numbers are said to be co-prime (or relatively prime) if their HCF is 1. They don’t need to be prime numbers themselves, just have no common factors other than 1. Example: 8 and 15 are co-prime.

2. Key Tricks & Shortcuts

Section titled “2. Key Tricks & Shortcuts”

  • Trick 1: Prime Factorization Method (HCF & LCM)

    • How to use it: Break down each number into its prime factors.
      • HCF: Take the lowest power of each common prime factor.
      • LCM: Take the highest power of all prime factors present in any of the numbers.
    • Example: Find the HCF and LCM of 24 and 36.
      • 24 = 23 x 31
      • 36 = 22 x 32
      • HCF = 22 x 31 = 12
      • LCM = 23 x 32 = 72

  • Trick 2: Division Method (HCF)

    • How to use it: Repeatedly divide the larger number by the smaller number, and then replace the larger number with the remainder until the remainder is 0. The last non-zero divisor is the HCF.
    • Example: Find the HCF of 48 and 72.
      • 72 ÷ 48 = 1 (remainder 24)
      • 48 ÷ 24 = 2 (remainder 0)
      • HCF = 24

  • Trick 3: Formula: HCF x LCM = Product of Two Numbers

    • How to use it: If you know the HCF and LCM of two numbers, you can find their product. Conversely, if you know the product and one of HCF or LCM, you can find the other.
    • Example: The HCF of two numbers is 12, and their LCM is 72. One of the numbers is 24. Find the other number.
      • Let the other number be x.
      • 12 x 72 = 24 x x
      • x = (12 x 72) / 24 = 36

  • Trick 4: Finding HCF and LCM of Fractions

    • How to use it:
      • HCF of Fractions = (HCF of Numerators) / (LCM of Denominators)
      • LCM of Fractions = (LCM of Numerators) / (HCF of Denominators)
    • Example: Find the HCF and LCM of 2/3, 4/5, and 6/7.
      • HCF = HCF(2, 4, 6) / LCM(3, 5, 7) = 2 / 105
      • LCM = LCM(2, 4, 6) / HCF(3, 5, 7) = 12 / 1 = 12

  • Trick 5: Handling Remainders (HCF)

    • How to use it: If you need to find the largest number that divides a, b, and c leaving the same remainder r in each case, then find the HCF of (a-r), (b-r), and (c-r).
    • Example: Find the largest number that divides 70, 105, and 175, leaving the same remainder in each case. If we subtract the smallest number from the other two, the difference will be a multiple of the divisor. So (105-70) = 35 and (175-105) = 70 and (175-70) = 105. Now find the HCF of 35, 70 and 105 which is 35. Thus 35 divides 70, 105 and 175 leaving remainder 0 in each case.

  • Trick 6: Handling Remainders (LCM)

    • How to use it: If you need to find the smallest number which when divided by a, b, and c leaves the same remainder r in each case, then the number is (LCM of a, b, c) + r.
    • Example: Find the smallest number which when divided by 12, 15, and 18 leaves a remainder of 5 in each case.
      • LCM of 12, 15, and 18 is 180.
      • Required number = 180 + 5 = 185

  • Trick 7: Vedic Math - Osculation Method (for Divisibility)

    • How to use it: This is particularly useful for checking the divisibility of large numbers by prime numbers.
      • For a divisor ending in 1 (e.g., 11, 31, 41): Double the last digit and subtract it from the remaining truncated number. Repeat until you get a small number divisible by the divisor.
      • For a divisor ending in 9 (e.g., 9, 19, 29): Multiply the last digit by the ‘osculator’ (which is the number you need to add to the last digit to get a multiple of 10). Add this to the truncated number. Repeat until you get a small number divisible by the divisor. For 19, the osculator is 2 (9 + (1 * 2) = 11, 11 + (1 * 2) = 13, 13 + (1 * 2) = 15, 15 + (1 * 2) = 17, 17 + (1 * 2) = 19). For 29, the osculator is 3.
    • Example: Check if 2147 is divisible by 11.
      • Truncated number: 214, Last digit: 7
      • 214 - (2 * 7) = 214 - 14 = 200
      • 20 - (2 * 0) = 20
      • 20 is not divisible by 11, so 2147 is not divisible by 11.

  • Trick 8: Cyclicity (for Unit Digits)

    • How to use it: Understand the repeating pattern of unit digits for powers of numbers. For example:
      • 2: 2, 4, 8, 6, 2, 4, 8, 6… (cycle of 4)
      • 3: 3, 9, 7, 1, 3, 9, 7, 1… (cycle of 4)
      • 4: 4, 6, 4, 6… (cycle of 2)
      • 5: 5, 5, 5, 5… (cycle of 1)
      • 6: 6, 6, 6, 6… (cycle of 1)
      • 7: 7, 9, 3, 1, 7, 9, 3, 1… (cycle of 4)
      • 8: 8, 4, 2, 6, 8, 4, 2, 6… (cycle of 4)
      • 9: 9, 1, 9, 1… (cycle of 2)
    • Example: What is the unit digit of 725?
      • The cycle of 7 is 4.
      • 25 divided by 4 gives a remainder of 1.
      • Therefore, the unit digit of 725 is the same as the unit digit of 71, which is 7.

  • Trick 9: Using Answer Choices (Problem Solving)

    • How to use it: When faced with HCF/LCM problems, especially those involving word problems, try substituting the answer choices back into the problem statement to see which one satisfies the given conditions. This can be much faster than solving the problem algebraically.

3. Essential Formulas & Rules

Section titled “3. Essential Formulas & Rules”
Formula/RuleDescription
HCF x LCM = Product of Two NumbersThis holds true only for two numbers.
HCF of Fractions = HCF(Numerators) / LCM(Denominators)Used to find the HCF of fractions.
LCM of Fractions = LCM(Numerators) / HCF(Denominators)Used to find the LCM of fractions.
Number = (Divisor x Quotient) + RemainderFundamental formula relating divisor, quotient, and remainder.
Sum of first n natural numbersn(n+1)/2
Sum of squares of first n natural numbersn(n+1)(2n+1)/6
Sum of cubes of first n natural numbers[n(n+1)/2]2
(a + b)2 = a2 + b2 + 2abA fundamental algebraic identity. Useful for simplifying expressions or solving problems.
(a - b)2 = a2 + b2 - 2abA fundamental algebraic identity. Useful for simplifying expressions or solving problems.
a2 - b2 = (a + b)(a - b)A fundamental algebraic identity. Useful for simplifying expressions or solving problems.
a3 + b3 = (a + b)(a2 - ab + b2)A fundamental algebraic identity. Useful for simplifying expressions or solving problems.
a3 - b3 = (a - b)(a2 + ab + b2)A fundamental algebraic identity. Useful for simplifying expressions or solving problems.

4. Detailed Solved Examples

Section titled “4. Detailed Solved Examples”

Example 1: [Basic]

Find the HCF and LCM of 12, 15, and 18 using the prime factorization method.

Solution:

  1. Prime factorization:
    • 12 = 22 x 31
    • 15 = 31 x 51
    • 18 = 21 x 32
  2. HCF: Take the lowest power of each common prime factor. The only common prime factor is 3. Lowest power is 31. Therefore, HCF = 3.
  3. LCM: Take the highest power of all prime factors present.
    • 22, 32, 51
    • LCM = 22 x 32 x 51 = 4 x 9 x 5 = 180

Example 2: [Medium]

The product of two numbers is 2028, and their HCF is 13. Find their LCM.

Solution:

  1. Apply the formula: HCF x LCM = Product of Two Numbers
  2. 13 x LCM = 2028
  3. LCM = 2028 / 13 = 156

Example 3: [Hard]

Find the greatest number that will divide 410, 751, and 1030 so as to leave the same remainder in each case.

Solution:

  1. Apply Trick 5 (Handling Remainders - HCF).
  2. Find the difference between the numbers:
    • 751 - 410 = 341
    • 1030 - 751 = 279
    • 1030 - 410 = 620
  3. Find the HCF of 341, 279, and 620.
    • Prime factorization: 341 = 11 x 31, 279 = 3 x 3 x 31, 620 = 2 x 2 x 5 x 31
    • HCF = 31
  4. Therefore, the greatest number that will divide 410, 751, and 1030 so as to leave the same remainder in each case is 31. (Note: 410/31 = 13 remainder 7, 751/31 = 24 remainder 7, 1030/31 = 33 remainder 7).

Example 4: [Medium]

Find the least number which when divided by 6, 9, 12, 15 leaves the same remainder 2 in each case.

Solution:

  1. Apply Trick 6 (Handling Remainders - LCM).
  2. Find the LCM of 6, 9, 12, and 15.
    • 6 = 2 x 3
    • 9 = 3 x 3
    • 12 = 2 x 2 x 3
    • 15 = 3 x 5
    • LCM = 2 x 2 x 3 x 3 x 5 = 180
  3. Add the remainder: 180 + 2 = 182

Therefore, the least number is 182.

5. Practice Problems (Graded Difficulty)

Section titled “5. Practice Problems (Graded Difficulty)”

[Easy] What is the HCF of 24 and 36?

[Easy] What is the LCM of 8 and 12?

[Medium] The HCF of two numbers is 8, and their LCM is 144. If one of the numbers is 16, find the other number.

[Medium] Find the largest number that will divide 122 and 243, leaving remainders of 2 and 3 respectively.

[Hard] Find the least number which when divided by 20, 25, 35 and 40 leaves remainders 14, 19, 29 and 34 respectively.

[Hard] The traffic lights at three different road crossings change after every 48 seconds, 72 seconds, and 108 seconds respectively. If they all change simultaneously at 8:20:00 hours, then at what time will they again change simultaneously?

6. Advanced/Case-Based Question

Section titled “6. Advanced/Case-Based Question”

A farmer has a field that is 288 meters long and 216 meters wide. He wants to divide the field into square plots of equal size.

a) What is the maximum possible size of each square plot?

b) How many such square plots can be created?

c) If the farmer wants to fence each plot at a cost of $5 per meter, what will be the total cost of fencing all the plots?