Govt. Exams
Entrance Exams
96 = 2⁵ × 3, 120 = 2³ × 3 × 5.
Common prime factors: 2³ × 3 = 8 × 3 = 24.
Consecutive even numbers differ by 2.
Examples: HCF(2,4)=2, HCF(6,8)=2, HCF(10,12)=2.
The HCF is always 2.
12 = 2² × 3, 15 = 3 × 5, 20 = 2² × 5. LCM = 2² × 3 × 5 = 60.
This is the smallest number divisible by all three.
For co-prime numbers, HCF = 1.
Using HCF × LCM = Product: 1 × 221 = Product.
Therefore, product = 221.
20 = 2² × 5, 30 = 2 × 3 × 5. LCM = 2² × 3 × 5 = 4 × 3 × 5 = 60.
Factors of 18: 1, 2, 3, 6, 9, 18.
Factors of 27: 1, 3, 9, 27.
Common factors: 1, 3, 9.
Highest common factor = 9.
Using the formula: HCF × LCM = Product of two numbers.
Therefore, 12 × LCM = 2160, so LCM = 2160 ÷ 12 = 180.
This question asks us to find the Least Common Multiple (LCM) of two numbers using prime factorization.
Break 15 into its prime factors.
Break 25 into its prime factors.
The LCM is found by taking the highest power of each prime that appears in either factorization: 3¹ and 5².
The LCM of 15 and 25 is 75.
This question asks us to find the Highest Common Factor (HCF) of two numbers using prime factorization or the Euclidean algorithm.
Express 56 as a product of prime numbers.
Express 72 as a product of prime numbers.
The HCF is the product of common prime factors with their lowest powers.
The HCF of 56 and 72 is 8, making the correct answer (B).
Prime factorization: 12 = 2² × 3, 18 = 2 × 3². LCM = 2² × 3² = 4 × 9 = 36 (taking highest powers of all prime factors).
