When working with numbers, especially in problems involving fractions, ratios, or divisibility, it is often useful to find the largest number that divides two or more integers exactly - without leaving any remainder. This number is called the Greatest Common Divisor, or GCD.
For example, if you want to simplify the fraction \(\frac{48}{180}\), you need to find the greatest number that divides both 48 and 180. This helps reduce the fraction to its simplest form. GCD also plays a crucial role in solving equations where integers are involved, such as Diophantine equations, and is a frequent topic in competitive exams.
In this chapter, we will explore what GCD is, how to find it using different methods, and how to apply it effectively in problem-solving.
The Greatest Common Divisor (GCD) of two integers \(a\) and \(b\) (not both zero) is the largest positive integer that divides both \(a\) and \(b\) without leaving a remainder.
Mathematically,
For example, consider the numbers 12 and 18:
The common factors are 1, 2, 3, and 6. The greatest among these is 6, so GCD(12, 18) = 6.
One straightforward way to find the GCD of two numbers is to break each number down into its prime factors and then identify the common prime factors with the smallest exponents.
Recall that a prime number is a number greater than 1 that has no positive divisors other than 1 and itself. Every number can be expressed as a product of primes, called its prime factorization.
Steps to find GCD by prime factorization:
From the factor trees:
Common prime factors are 2 and 3.
Therefore,
\[\mathrm{GCD}(48,180) = 2^2 \times 3^1 = 4 \times 3 = 12\]While prime factorization is straightforward, it can be time-consuming for large numbers. The Euclidean algorithm is a faster, more efficient method to find the GCD using repeated division.
The key idea is based on the fact that the GCD of two numbers also divides their difference.
Given two integers \(a\) and \(b\) where \(a > b\), the Euclidean algorithm uses the following recursive relation:
This means you replace the pair \((a,b)\) with \((b, a \bmod b)\) and repeat until the remainder is zero. The divisor at that step is the GCD.
graph TD A[Start with numbers a and b (a > b)] B[Divide a by b to get remainder r] C{Is r = 0?} D[Yes: GCD is b] E[No: Replace a with b, b with r] F[Repeat division] A --> B --> C C -->|Yes| D C -->|No| E --> F --> CStep 1: Find prime factors of 48.
48 = 2 x 24 = 2 x 2 x 12 = 2 x 2 x 2 x 6 = 2 x 2 x 2 x 2 x 3 = \(2^4 \times 3\)
Step 2: Find prime factors of 180.
180 = 2 x 90 = 2 x 2 x 45 = 2 x 2 x 3 x 15 = 2 x 2 x 3 x 3 x 5 = \(2^2 \times 3^2 \times 5\)
Step 3: Identify common prime factors with minimum powers.
Step 4: Multiply these to get GCD.
\(\mathrm{GCD} = 2^2 \times 3 = 4 \times 3 = 12\)
Answer: The GCD of 48 and 180 is 12.
Step 1: Divide 252 by 105 and find the remainder.
252 / 105 = 2 remainder 42 (since \(252 = 105 \times 2 + 42\))
Step 2: Replace (252, 105) with (105, 42).
Step 3: Divide 105 by 42.
105 / 42 = 2 remainder 21 (since \(105 = 42 \times 2 + 21\))
Step 4: Replace (105, 42) with (42, 21).
Step 5: Divide 42 by 21.
42 / 21 = 2 remainder 0 (since \(42 = 21 \times 2 + 0\))
Step 6: Since remainder is 0, GCD is 21.
Answer: The GCD of 252 and 105 is 21.
Step 1: Find GCD of numerator and denominator.
Use Euclidean algorithm:
770 / 462 = 1 remainder 308
462 / 308 = 1 remainder 154
308 / 154 = 2 remainder 0
GCD = 154
Step 2: Divide numerator and denominator by GCD.
\(\frac{462}{770} = \frac{462 \div 154}{770 \div 154} = \frac{3}{5}\)
Answer: The simplified fraction is \(\frac{3}{5}\).
Step 1: Find the GCD of 360 and 480.
Using Euclidean algorithm:
480 / 360 = 1 remainder 120
360 / 120 = 3 remainder 0
GCD = 120
Step 2: The maximum value of each part is INR 120.
Step 3: Number of parts for each amount:
Answer: The amounts can be divided into parts of INR 120 each.
Step 1: Recall the relation between GCD and LCM:
Step 2: Substitute known values:
\[ 12 \times 180 = 36 \times b \] \[ 2160 = 36 \times b \]Step 3: Solve for \(b\):
\[ b = \frac{2160}{36} = 60 \]Answer: The other number is 60.
When to use: When numbers are large and prime factorization is time-consuming.
When to use: When one of the numbers is zero, this shortcut quickly gives the GCD.
When to use: When either GCD or LCM is missing in a problem.
When to use: Before factorization to save time and avoid errors.
When to use: When asked to reduce fractions to their lowest terms.
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