Problems on H.C.F and L.C.M

Introduction

The Highest Common Factor (H.C.F) and Least Common Multiple (L.C.M) are fundamental concepts in elementary mathematics. They have widespread applications across a range of problems in the quantitative aptitude section of competitive examinations like SSC (Staff Selection Commission), RRB (Railway Recruitment Board), and various banking exams.

H.C.F, also known as the Greatest Common Divisor (GCD), is the largest number that perfectly divides a set of numbers. Conversely, the L.C.M is the smallest number that is a multiple of each number in a set.

This chapter aims to provide a comprehensive understanding of these concepts, their applications, formulas, and associated problems, punctuated by a host of examples to ensure a firm grasp of the topic.

H.C.F and L.C.M in Competitive Exams

Questions based on H.C.F and L.C.M appear frequently in the SSC, RRB, and banking exams. A firm understanding of these topics can help significantly increase your scores in these exams. Here is a breakdown of the importance of these topics across various exams:

1. SSC: H.C.F and L.C.M questions usually revolve around identifying the H.C.F or L.C.M of given numbers, or solving word problems related to these topics and the number system.
2. RRB: Questions on these topics often deal with time, speed, and distance, as well as problems on ages, work, and wages.
3. Banking Exams: H.C.F and L.C.M are extensively used in problems related to fractions, ratios, and percentages. They are also utilized in problems related to time and work, pipes and cisterns, and calendars.

Basic Formulas and Concepts

Before proceeding to problem-solving, it is crucial to understand the basic formulas and concepts related to H.C.F and L.C.M:

1. H.C.F (Highest Common Factor):
   • Euclidean Algorithm: Divide ‘a’ by ‘b’ and get the remainder ‘r’. Replace ‘a’ with ‘b’ and ‘b’ with ‘r’ and repeat the process until ‘b’ is zero. The H.C.F will be the last non-zero remainder.
   Example: Find the H.C.F of 48 and 18.
   • Divide 48 by 18 to get a quotient of 2 and a remainder of 12.
   • Now, replace 48 with 18 and 18 with 12, and repeat the process. The remainder now is 6.
   • Repeat the process one more time. Now, the remainder is zero, which means our H.C.F is the last non-zero remainder, 6.
2. L.C.M (Least Common Multiple):
   • Formula: If you have two numbers ‘a’ and ‘b’, their L.C.M is calculated as: L.C.M = (a*b)/H.C.F(a,b). Note: You need to find the H.C.F first to find the L.C.M using this formula.
   Example: Using the H.C.F from the previous example, let’s find the L.C.M of 48 and 18.
   • Apply the formula, L.C.M = (ab)/H.C.F. = (4818)/6 = 144.
3. Properties of H.C.F and L.C.M:
   • For any two numbers ‘a’ and ‘b’, the product of the H.C.F and L.C.M is equal to the product of these two numbers: H.C.F(a, b) * L.C.M(a, b) = a * b.
   • The H.C.F of fractions = H.C.F of Numerators/L.C.M of Denominators.
   • The L.C.M of fractions = L.C.M of Numerators/H.C.F of Denominators.

Advanced Concepts and Problem Types

Now that we’ve covered the basics let’s look at some more complex concepts and problem types related to H.C.F and L.C.M:

1. Co-Primes: Two numbers are said to be co-prime or mutually prime if their H.C.F is 1. They do not have any common factor other than 1.
2. Problems Related to Numbers: These can include finding the L.C.M or H.C.F of given numbers or finding numbers given their L.C.M and H.C.F.Example: If the H.C.F and L.C.M of two numbers are 5 and 120 respectively, and one of the numbers is 20, find the other number.
   • Using the property H.C.F * L.C.M = Product of the numbers, we can set up the equation 5 * 120 = 20 * x, where x is the other number. Solving this equation gives us x = 30.
3. Word Problems: These problems apply the concepts of L.C.M and H.C.F to practical situations, like determining the least amount of time required for two events with different periods to coincide.Example: Two traffic lights at different points start flashing at 6 A.M. One flashes every 4 seconds and the other every 6 seconds. When will they flash together next?
   • This problem is asking for the L.C.M of 4 and 6 seconds. The L.C.M of 4 and 6 is 12. Therefore, the two lights will flash together after 12 seconds.
4. Problems Involving Ratio or Fractions: Since the H.C.F and L.C.M of fractions involve both the numerators and the denominators, problems involving the comparison of different fractions often appear in competitive exams.Example: If the H.C.F of two fractions 3/4 and 5/2 is 1/4, find their L.C.M.
   • The L.C.M of fractions = L.C.M of Numerators/H.C.F of Denominators. The L.C.M of 3 and 5 is 15, so the L.C.M of the fractions is 15/4.