Finding the H.C.F. of Three Numbers

The Highest Common Factor (H.C.F.) of three numbers is the largest number that divides all three numbers exactly, leaving no remainder. Understanding H.C.F. is essential in various mathematical applications, including simplifying fractions, solving problems involving ratios, and understanding the relationships between numbers.

Methods to Find H.C.F.

There are several methods to find the H.C.F. of three numbers. Let’s explore the most common ones:

1. Prime Factorization Method

This method involves breaking down each number into its prime factors. The prime factors are the building blocks of a number, and they are prime numbers themselves (numbers greater than 1 that are only divisible by 1 and themselves, like 2, 3, 5, 7, 11, and so on).

Steps:

  1. Prime Factorization: Find the prime factors of each of the three numbers.
  2. Common Factors: Identify the prime factors that are common to all three numbers.
  3. H.C.F.: Multiply the common prime factors together. The result is the H.C.F.

Example:

Let’s find the H.C.F. of 12, 18, and 24.

  1. Prime Factorization:
    • 12 = 2 x 2 x 3
    • 18 = 2 x 3 x 3
    • 24 = 2 x 2 x 2 x 3
  2. Common Factors: The common prime factors are 2 and 3.
  3. H.C.F.: 2 x 3 = 6. Therefore, the H.C.F. of 12, 18, and 24 is 6.

2. Division Method

This method involves repeatedly dividing the larger number by the smaller number until a remainder of 0 is obtained. The last divisor is the H.C.F.

Steps:

  1. Divide: Divide the largest number by the smallest number.
  2. Replace: Replace the larger number with the smaller number and the smaller number with the remainder.
  3. Repeat: Repeat steps 1 and 2 until the remainder is 0.
  4. H.C.F.: The last divisor is the H.C.F.

Example:

Let’s find the H.C.F. of 36, 48, and 60.

  1. Divide: 60 ÷ 36 = 1 (remainder 24)
  2. Replace: 36 becomes 24 and 24 becomes 12.
  3. Repeat: 24 ÷ 12 = 2 (remainder 0)
  4. H.C.F.: The last divisor is 12. Therefore, the H.C.F. of 36, 48, and 60 is 12.

3. Listing Factors Method

This method involves listing all the factors of each number and then identifying the common factors. The largest common factor is the H.C.F.

Steps:

  1. Factors: List all the factors of each of the three numbers.
  2. Common Factors: Identify the factors that are common to all three numbers.
  3. H.C.F.: The largest common factor is the H.C.F.

Example:

Let’s find the H.C.F. of 15, 20, and 25.

  1. Factors:
    • Factors of 15: 1, 3, 5, 15
    • Factors of 20: 1, 2, 4, 5, 10, 20
    • Factors of 25: 1, 5, 25
  2. Common Factors: The common factors are 1 and 5.
  3. H.C.F.: The largest common factor is 5. Therefore, the H.C.F. of 15, 20, and 25 is 5.

Applications of H.C.F.

The H.C.F. has several practical applications in mathematics and real-life scenarios:

1. Simplifying Fractions

The H.C.F. is used to simplify fractions to their simplest form. To simplify a fraction, we divide both the numerator and denominator by their H.C.F.

Example:

Simplify the fraction 12/18.

The H.C.F. of 12 and 18 is 6. Dividing both the numerator and denominator by 6, we get:

12 ÷ 6 / 18 ÷ 6 = 2/3

2. Solving Ratio Problems

The H.C.F. can be used to simplify ratios. Ratios represent the relative sizes of two or more quantities. By dividing all the terms of a ratio by their H.C.F., we can simplify the ratio to its simplest form.

Example:

Simplify the ratio 12:18:24.

The H.C.F. of 12, 18, and 24 is 6. Dividing each term of the ratio by 6, we get:

12 ÷ 6 : 18 ÷ 6 : 24 ÷ 6 = 2:3:4

3. Sharing Equally

The H.C.F. can be used to determine the largest number of equal groups that can be formed from a given set of items.

Example:

You have 12 apples, 18 oranges, and 24 pears. What is the largest number of identical fruit baskets you can make using all the fruits?

The H.C.F. of 12, 18, and 24 is 6. Therefore, you can make 6 fruit baskets, each containing 2 apples, 3 oranges, and 4 pears.

Conclusion

Finding the H.C.F. of three numbers is a fundamental skill in mathematics. Understanding the different methods and their applications allows us to solve various problems involving numbers, fractions, ratios, and real-life scenarios.

Citations

  1. 1. Math is Fun – HCF
  2. 2. Khan Academy – Greatest Common Factor
  3. 3. Cuemath – HCF

Related

(2) O3 + H → O2 + OH k2 = 1.78×10^-11 cm^3 s^-1 (3) O + OH → O2 + H k3 = 4.40×10^-11 cm^3 s^-1 (5) O + HO2 → O2 + OH k5 = 3.50×10^-11 cm^3 s^-1 (6) H + HO2 → O2 + H2 k6 = 5.40×10^-12 cm^3 s^-1 (9) OH + HO2 → O2 + H2O2 k9 = 4.00×10^-11 cm^3 s^-1 (10) HO2 + HO2 → O2 + H2O2 k10 = 2.50×10^-12 cm s^-1 (11) O + O2 + M → O3 + M k11 = 1.05×10^-34 cm^6 s^-1 (14) H + O2 + M → HO2 + M k14 = 8.08×10^-32 cm^6 s^-1 (15) H + H + M → H2O + M k15 = 3.31×10^-27 cm^6 s^-1 (16) O2 + hv → 2 O k16 = (1.26×10^-8 s^-1) φ (17) H2O + hv → H + OH k17 = (3.4×10^-6 s^-1) φ (18) O3 + hv → O2 + O k18 = (7.10×10^-5 s^-1) φ

Table 1 Reactions, rate constants and activation energies used in the model* No. Reaction kopt (M⁻¹ s⁻¹) 1 OH + H₂ → H + H₂O 3.74 x 10⁷ 2 OH + HO₂ → HO₂ + OH⁻ 5 x 10⁹ 3 OH + H₂O₂ → HO₂ + H₂O 3.8 x 10⁷ 4 OH + O₂ → O₂ + OH 9.96 x 10⁹ 5 OH + HO₂ → O₂ + H₂O 7.1 x 10⁹ 6 OH + OH → H₂O₂ 5.3 x 10⁹ 7 OH + e⁻aq → OH⁻ 3 x 10¹⁰ 8 H + O₂ → HO₂ 2.0 x 10¹⁰ 9 H + HO₂ → H₂O₂ 2.0 x 10¹⁰ 10 H + H₂O₂ → OH + H₂O 3.44 x 10⁷ 11 H + OH → H₂O 1.4 x 10¹⁰ 12 H + H → H₂ 1.94 x 10¹⁰ 13 e⁻aq + O₂ → O₂⁻ 1.9 x 10¹⁰ 14 e⁻aq + O₂ → HO₂⁻ + OH⁻ 1.3 x 10¹⁰ 15 e⁻aq + HO₂ 2.0 x 10¹⁰ 16 e⁻aq + H₂O₂ 1.1 x 10¹⁰ 17 e⁻aq + HO₂ → OH + OH⁻ 1.3 x 10¹⁰ 18 e⁻aq + H⁺ → H 2.3 x 10¹⁰ 19 e⁻aq + e⁻aq → H₂ + OH⁻ + OH⁻ 2.5 x 10⁹ 20 HO₂ + O₂ → O₂ + HO₂ 1.3 x 10⁹ 21 HO₂ + HO₂ → O₂ + H₂O₂ 8.3 x 10⁵ 22 HO₂ + HO₂ → O₂ + OH + H₂O 3.7 23 HO₂ + HO₂ → O₂ + O₂ + OH + H₂O 7 x 10⁵ s⁻¹ 24 H⁺ + O₂⁻ → HO₂ 4.5 x 10¹⁰ 25 H⁺ + O₂⁻ → O₂ 2.0 x 10¹⁰ 26 H⁺ + OH⁻ 1.4 x 10¹¹ 27 H⁺ + HO₂⁻ 2 x 10¹⁰ 28 H₂O₂ → HO₂ + H⁺ + OH⁻ 2.5 x 10⁻⁵ s⁻¹ 29 H₂O₂ → H⁺ + OH⁻ 1.4 x 10⁻⁷ s⁻¹ 30 O₂ + O₂ → O₂ + HO₂ + OH⁻ 0.3 31 O₂ + H₂O₂ → O₂ + OH + OH 16 32

(2) O3 + H → O2 + OH k2 = 1.78×10^-11 cm^3 s^-1 (3) O + OH → O2 + H k3 = 4.40×10^-11 cm^3 s^-1 (5) O + HO2 → O2 + OH k5 = 3.50×10^-11 cm^3 s^-1 (6) H2O + O → 2 OH k6 = 5.40×10^-12 cm^3 s^-1 (9) OH + HO2 → O2 + H2O k9 = 4.00×10^-11 cm^3 s^-1 (10) HO2 + HO2 → O2 + H2O2 k10 = 2.50×10^-12 cm s^-1 (11) O + O2 + M → O3 + M k11 = 1.05×10^-34 cm^6 s^-1 (14) H + O2 + M → HO2 + M k14 = 8.08×10^-32 cm^6 s^-1 (15) OH + H + M → H2O + M k15 = 3.31×10^-27 cm^6 s^-1 (16) O2 + hv → 2 O k16 = (1.26×10^-8 s^-1) φ (17) H2O + hv → H + OH k17 = (3.4×10^-6 s^-1) φ (18) O3 + hv → O2 + O k18 = (7.10×10^-8 s^-1) φ