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**Learning Objectives**

**HCF & LCM**

- Definitions
- Methods & Steps of solving Problems
- Shortcut techniques for saving time in solving problems

**Introduction**

The chapter of HCF & LCM is part of Number Systems unit which is amongst the **most important chapters** in the entire syllabus of Quantitative Aptitude for the CAT examination (and also for other parallel MBA entrance exams). Students are advised to go through this chapter with utmost care understanding each concept and question type on this topic. The CAT has consistently contained anything between **20-40% of the marks** based on questions taken from this chapter.

Naturally, this chapter becomes one of the most crucial as far as your quest to reach close to the qualification score in the section of Quantitative Aptitude and Data Interpretation is concerned.

**Factors and Multiples**

If number a divided another number b exactly, we say that a is a factor of b. In this case, b is called a multiple of a.

**Highest Common Factor (H.C.F.) **

- H.C.F is also called as
**Greatest Common Measure**(G.C.M.) or**Greatest Common Divisor**(G.C.D.) - Consider two natural numbers E and R. If the numbers E and R are exactly divisible by the same number x, then x is a common divisor of E and R. The highest of all the common divisors of E and R is called as the GCD or the HCF. This is denoted as GCD (E, R).
- The H.C.F. of two or more than two numbers is the greatest number that divides each of them exactly.

**➤ Finding the GCD / H.C.F. of two numbers E and R**

- Find the standard form of the numbers E and R.
- Write out all prime factors that are common to the standard forms of the numbers E and R.
- Raise each of the common prime factors listed above to the lesser of the powers in which it appears in the standard forms of the numbers E and R.
- The product of the results of the previous step will be the GCD of E and R.
- H.C.F of (4,6):

**Illustration:** Find the GCD of 150, 210, 375.

**Step 1:**Writing down the standard form of numbers

► 150 = 5 X 5 X 3 X 2

► 210 = 5 X 2 X 7 X 3

► 375 = 5 X 5 X 5 X 3**Step 2:**Writing Prime factors common to all the three numbers is 5 X 3.**Step 3:**This will give the same result, i.e. 5 X 3.**Step 4:**Hence, the HCF will be 5 X 3 = 15.

__There are two methods of finding the H.C.F. of a given set of numbers:__

**(i) Factorization Method **

- Also called as
**Prime Factorisation Method.** - Express each one of the given numbers as the product of prime factors.
- The product of the least powers of common prime factors gives H.C.F.

**» Stepwise Explanation of Solving the Problem**

**Step 1:**Express each of the given numbers as the product of prime factors.**Step 2:**The product of terms containing the least powers of common prime factors gives the H.C.F. of the given numbers.

**(ii) Division Method**

- Suppose we have to find the H.C.F. of two given numbers, divide the larger by, the smaller one.
- Now, divide the divisor by the remainder.
- Repeat the process of dividing the preceding number by the remainder last obtained till zero is obtained as remainder.
- The last divisor is required H.C.F.

**» ****Stepwise Explanation of Solving the Problem**

**Step 1:**Divide the larger number by the smaller number and obtain the remainder.**Step 2:**Divide the divisor by the remainder obtained in Step 1.**Step 3:**Repeat the process of dividing the preceding divisor by the remainder last obtained till you get 0 as remainder.**Step 4:**The last divisor is the required H.C.F.

**Example: Find the L.C.M. of 72, 240, 196.****Solution: ****(i) Using Prime factorisation method:**

► 72 = 2 × 2 × 2 × 3 × 3 = 2^{3 }× 3^{2}

► 240 = 2 × 2 × 2 × 2 × 3 × 5 = 2^{4 }× 3 × 5

► 196 = 2 × 2 × 7 × 7 = 2^{2 }× 7^{2}

L.C.M. of the given numbers = Product of all the prime factors of each of the given number with greatest index of common prime factors

= 2^{4 }× 3^{2 }× 5 × 7^{2} = 16 × 9 × 5 × 49 = 35280.

**(ii) Using the Division method:**__2 | 72, 240, 196____2 | 36, 120, 98____2 | 18, 60 , 49____3 | 9 , 30 , 49__

L.C.M. of the given numbers

= Product of divisors and the remaining numbers

= 2 × 2 × 2 × 3 × 3 × 10 × 49

= 35280

**➤ **

The CAT exam is all about saving time, hence we suggest you keenly observe & practice the examples on these short tricks to solve the problems, as they save your time in the final exam.

- The above mentioned illustration is however extremely cumbersome and time taking. Let us look at a much faster way of finding the HCF of a set of numbers.
**Suppose you were required to find the HCF of 39,78 and 195.****Logic**: The HCF of these numbers would necessarily have to be a factor (divisor) of the difference between any pair of numbers from the above 3. i.e. the HCF has to be a factor of (78 – 39 = 39) as well as of (195 – 39 = 156) and (195 – 78 = 117). Why? Well, the logic is simple if you were to consider the tables of numbers on the number line.- For any two numbers on the number line, a common divisor would be one which divides both. However, for any number to be able to divide both the numbers, it can only do so if it is a factor of the difference between the two numbers. Got it?? No??
__Then, Let’s see an illustrative:__Say we take the numbers 68 and 119. The difference between them being 51, it is not possible for any number outside the factor list of 51 to divide both 68 and 119. Thus, for example, a number like 4, which divides 68 can never divide any number which is 51 away from 68 because 4 is not a factor of 51. Only factors of 51, i.e. 51,17,3 and 1 ‘could’ divide both these numbers simultaneously.

**Example 1: The sides of a hexagonal field are 216, 423, 1215, 1422, 2169 and 2223 metres. Find the greatest length of tape that would be able to exactly measure each of these sides without having to use fractions/parts of the tape?****Solution: **Before solving this example, remind yourself with the above short trick you read above.

► In this question, we are required to identify the HCF of the numbers 216, 423,1215, 1422, 2169 and 2223.

► In order to do that, we first find the smallest difference between any two of these numbers. It can be seen that the difference between 2223-2169 = 54.

Thus, the required HCF would be a factor of the number 54. The factors of 54 are: 1 × 54 2 × 27 3 X 18 6 X 9

► One of these 8 numbers has to be the HCF of the 6 numbers. 54 cannot be the HCF because the numbers 423 and 2223 being odd numbers would not be divisible by any even number. Thus, we do not need to check any even numbers in the list.

► 27 does not divide 423 and hence cannot be the HCF. 18 can be skipped as it is even.

**► Checking for 9:** 9 divides 216,423,1215,1422 and 2169.

► Hence, it would become the HCF. (Note: we do not need to check 2223 once we know that 2169 is divisible by 9)

Question 1:A nursery has 363,429 and 693 plants respectively of 3 distinct varieties. It is desired to place these plants in straight rows of plants of 1 variety only so that the number of rows required is the minimum. What is the size of each row and how many rows would be required? (Try to solve using the shortcut method)?

**Least Common Multiple (L.C.M.)**

- Let E and R be two natural numbers distinct from each other. The smallest natural number n that is exactly divisible by E and R is called the Least Common Multiple (LCM) of E and R and is designated as LCM (E, R).

The least number which is exactly divisible by each one of the given numbers is called their L.C.M.

**➢ Finding the LCM two numbers E and R**

- Find the standard form of the numbers E and R.
- Write out all the prime factors, which are contained in the standard forms of either of the numbers.
- Raise each of the prime factors listed above to the highest of the powers in which it appears in the standard forms of the numbers E and R.
- The product of the results of the previous step will be the LCM of E and R.
- L.C.M of (4,6) is:

**Illustration:** Find the LCM of 150, 210, 375.

**Step 1: **Writing down the standard form of numbers:

► 150 = 5 × 5 × 3 × 2

► 210 = 5 × 2 × 7 × 3

► 375 = 5 × 5 × 5 × 3**Step 2: **Write down all the prime factors: that appear at least once in any of the numbers: 5, 3, 2, 7.**Step 3: **Raise each of the prime factors to their highest available power (considering each to the numbers).

The LCM = 2 × 3 × 5 × 5 × 5 × 7 = 5250.

Important Rule:GCD (E, R) × LCM (E, R) = E × R

i.e. The product of the HCF and the LCM equals the product of the numbers.Note:This rule is applicable only for two numbers

__There are two methods of finding the L.C.M. of a given set of numbers:__

**(i) Factorization Method**

Resolve each one of the given numbers into a product of prime factors. Then, L.C.M. is the product of the highest powers of all the factors.

**» Stepwise Explanation of Solving Problem**

**Step 1:**Express each of the given numbers as the product of prime factors.**Step 2:**The product of terms containing the highest powers of all the factors gives the L.C.M. of the given numbers.

**(ii) Division Method (short-cut)**

- Arrange the given numbers in a row in any order.
- Divide by a number which divided exactly at least two of the given numbers and carry forward the numbers which are not divisible.
- Repeat the above process till no two of the numbers are divisible by the same number except 1.
- The product of the divisors and the undivided numbers is the required L.C.M. of the given numbers.

**» ****Stepwise Explanation of Solving Problem**

**Step 1:**Arrange the given number in a row.**Step 2:**Divide by a number which divides exactly at least two of the given numbers and carry forward the numbers which are not divisible.**Step 3:**Repeat the above process till no two numbers are divisible by the same number except 1.**Step 4:**The product of the divisors and the undivided numbers is the required L.C.M. of the given numbers.

Question 2:The LCM of two numbers is 936. If their HCF is 4 and one of the numbers is 72, the other is:

Question 3:Find Greatest Number, which will divide 215,167 and 135 so as to leave the same remainder in each case

Co-primes:Two numbers are said to be co-primes if their H.C.F. is 1.

**H.C.F. and L.C.M. of Decimal Fractions**

In given numbers, make the same number of decimal places by annexing zeros in some numbers, if necessary. Considering these numbers without a decimal point, find H.C.F. or L.C.M. as the case may be. Now, in the result, mark off as many decimal places as are there in each of the given numbers.

**Comparison of Fractions**

Find the L.C.M. of the denominators of the given fractions. Convert each of the fractions into an equivalent fraction with L.C.M as the denominator, by multiplying both the numerator and denominator by the same number. The resultant fraction with the greatest numerator is the greatest.

Question 4:What Will Be The Least Possible Number Of The Planks, if three pieces of timber 42 m, 49 m, and 63 m long have to be divided into planks of the same length?

Next DocumentIn the next Document you will be practising CAT level & previous CAT, XAT problems to improve your understanding in the concept of HCF & LCM.

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