Properties of HCF (GCD) & LCM

To understand the concept of HCF (Highest Common Factor) / GCD (Greatest Common Divisor) & LCM (Lowest Common Multiple) in completely, we have to recall the terms Factors and Multiples. 

We may also recall Least Common Multiple ( LCM) and Highest Common Factor (HCF) or Greatest Common Divisor (GCD) before going through their properties.

Least Common Multiple ( LCM)

The least common multiple ( LCM ) is also referred to as the lowest common multiple or smallest common multiple of two integers. For any two integers a and b, usually denoted by LCM(a, b), is the smallest positive integer that is divisible by both a and b.

For example, The  L.C.M of 3 and 4 is 12.

Highest Common Factor (HCF) Or Greatest Common Divisor (GCD / gcd) 

The Highest Common Factor ( HCF) or greatest common divisor (GCD / gcd) of two or more integers, which are not all zero, is the largest positive integer that divides each of the integers.
For any two integers a and b, usually denoted by HCF (a, b), is the largest integer that can divide both  numbers a and b. (a , b) can be also written as for simplicity. 
For example, The  H.C.F of 10 and 15 is 5.

Important Properties of HCF and LCM

• 1) The Highest Common Factor ( HCF) or greatest common divisor (GCD) of two or more integers is always LESS  than  to the  given numbers. 

For example, The  H.C.F of 10 and 15 is 5.                                 

If we consider another example,The  H.C.F of 14 and 28 is 14.

• 2) The Least Common Multiple  LCM)  of two or more integers is always GREATER than  or EQUALS to  given numbers.                                                       

For example, The  L.C.M of 10 and 20 is 20.  

If we consider another example,The  L.C.M of 10, 15 and 20 is 60.

• 3)  Relation between LCM and HCF (Formula for finding HCF and LCM)

Given relations is very popular in mathematics to find either LCM (if HCF is known) or HCF (if LCM is known)  or both are known along with one number then we can find another unknown number.

'The product of Least Common Multiple( LCM) and Highest Common Factor (HCF) or Greatest Common Divisor (GCD) of any two Numbers/ Polynomials is equaled to their Products'. 

in other words, mathematically we can write as:-

                    LCM × HCF = Product of the Numbers

if a and b are two numbers, then.

                    HCF (a , b) x LCM (a , b)   = a × b.

Above results  can be used directly or summarized as in different situations

LCM (a , b)   = (a × b) / HCF (a , b).

HCF (a , b) = (a × b) / LCM (a , b).

a ( one number) = LCM (a , b) × HCF (a , b) / b 

Illustrative Examples for this third property / Formula / Results.

Example 1:- The LCM and HCF of two numbers are 150 and 4 respectively. If  25 is a number given then find the other number.

Solution: As we know  that the product of two numbers is equal to the product of HCF and LCM of two numbers & HCF and LCM are given to us. Also, one of the numbers is given to us. Thus, we need to find the other number so we may use the above relation.

So,  HCF x LCM (or LCM x HCF )  = Product of the Numbers

=> 150 × 4 = 25 × x

=> x = 24

Therefore  required answer is 24.

OR

We can use direct relation to verify the above results

a ( one number) = LCM (a , b) × HCF (a , b) / b 

a = (150 x 4 ) / 25

a = 24.

Therefore  required answer is 24.

Example 2:-  Prove the above formula 

 LCM (18 & 12) × HCF (18 & 12) = Product of 18 and 12

Solution:- First of all we have to find the prime factorisation of given Numbers 18 and 12

18 = 2 x3 x 3 =  2 x 3²

12 = 2 x 2 x 3 = 2² x 3 

Now we have to calculate their LCM and HCF

LCM of 18 and 12 = 2² × 3² = 4 × 9 = 36

HCF of 18 and 12 = 3 × 2 = 6

LHS (Left Hand Side) =

LCM (18 & 12) × HCF (18 & 12) = 36 × 6 = 216

RHS (Right Hand Side) = 

Product of 18 and 12 = 18 × 12 = 216

Now 

LHS (Left Hand Side) =  RHS (Right Hand Side) = 216

Hence, LCM (18 & 12) × HCF (18 & 12) =  18 × 12 = 216

• 4)  Formula for finding HCF and LCM of Fraction is such as 

L.C.M. = L.C.M. of Numerator / H.C.F. of Denominator

H.C.F. = H.C.F. of Numerator / L.C.M. of Denominator 

Now we solve some problems for finding LCM and HCF of Fractions:-

CFofnumeratorsLCMofdenominators

Example 3:-  Find L.C.M. of 

                     4 / 3, 8 / 81, 64 / 9 ,10 / 27 

Solution: 

Before using this formula we have  to find LCM of Numerator and HCF of Denominator.

L.C.M. =	L.C.M. of Numerator
	H.C.F. of Denominator

L.C.M. of Numerators = 4, 8, 64, 10 ( By Prime Factorisation)

4 = 2²
8 = 2³
64 = 2⁶
10 = 2 × 5
L.C.M of 4, 8, 64, 10 = 2⁶ × 5 = 320

H.C.F. of Denominators = 3, 81, 9, 27  ( By Prime Factorisation)
3 = 3¹
81 = 3⁴
9 = 3²
27 = 3³
H.C.F. of 3, 81, 9, 27 = 3

L.C.M. of 

                     4 / 3, 8 / 81, 64 / 9 ,10 / 27  = 320 / 3

Therefore 320/ 3 is our required Answer.

We will solve one more Examples

Example 4:- Find the HCF of 

415, 910, 1835, 2130

Solution: Before using this formula we  have to find LCM of Denominator and HCF of Numerator.

H.C.F. =	H.C.F. of Numerator
	L.C.M. of Denominator

H.C.F. of Numerators = 4, 9, 18, 21  ( By Prime Factorisation) 

4 = 2 × 2 
9 = 3 × 3
18 = 2 × 3 × 3
21 = 3 × 7
HCF (4, 9, 18, 21) = 1
L.C.M. of Denominators = 15,10,35, 30 ( By Prime Factorisation)

15 = 3 × 5
10 = 2 × 5
35 = 5 × 7
30 = 2 × 3 × 5
LCM(15, 10, 35, 30) = 5 × 3× 2 ×  7 = 210
The required 

HCF = HCF(4, 9, 18, 21)/LCM(4, 9, 18, 21) = 1/210

Therefore 1 / 210  is our required Answer.

Example 5:- Find the HCF of 

415, 1681, 89.

Solution: Before using this formula we have to find LCM of Denominator and HCF of Numerator.

H.C.F. =	H.C.F. of Numerator
	L.C.M. of Denominator

OR                H.C.F. =	H.C.F. of 4, 16, 8
	L.C.M. of 15, 81, 9

H.C.F. of Numerators = 4, 16, 8  ( By Prime Factorisation)

4= 2²
16 = 2⁴

8 = 2³
Number with least index = 2² = 4
H.C.F. of 4, 16, 8 = 4

L.C.M. of Denominators = 15,81,9 ( By Prime Factorisation)

15= 3 x 5¹
9 = 3²
81 = 3⁴
Number with highest index =  5 x 3⁴  = 5 x 81 =405
L.C.M. of 15, 81, 9 = 405

H.C.F. =	4
	405

Therefore 4 / 405  is our required Answer.

• 5 ) Since by the definition of the co-prime number we know that "HCF of co-prime numbers is 1". Therefore LCM of given co-prime numbers is equal to the product of the co-prime numbers.

Co-prime Numbers LCM  = Product Of The Numbers

Example 6:-  Prove the above formula 

 LCM of Co-prime Numbers 5&7 = Product Of The Numbers 5&7

Solution:- 

LHS (Left Hand Side) =
      5= 5¹                  7 = 7¹

LCM (5 & 7) = 5 × 7 = 35

RHS (Right Hand Side) = 

Product of 5 and 7 = 5× 7 = 35

Now 

LHS (Left Hand Side) =  RHS (Right Hand Side) = 35.

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Highest Common Factor or Greatest Common Divisor

Highest Common Factor (HCF)

or 

Greatest Common Divisor ( GCD)

In arithmetic ( mathematics) and number theory, the greatest common divisor (GCD/gcd) of two or more integers, which are not all zero, is the largest positive integer that divides each of the integers. The greatest number which divides each of the two or more numbers is popularly called HCF or Highest Common Factor (HCF). It is also called the Greatest Common Measure (GCM) and Greatest Common Divisor(GCD).
For any two integers a and b, usually denoted by HCF (a, b), is the largest integer that can divide both numbers a and b. (a , b)  can be also written as for simplicity. 
The HCF of more than two integers is also well-defined: it is the largest integer that can divide to all of them.

What is the Highest Common Factor (HCF)?
It is a method to find the highest common factor between numbers ( any two or more). Generally according to definitions the common factor is a number which is a factor of two or more numbers therefore HCF is used to determine the common factor. 
Example 1:-
If we consider an example, To find HCF of 2 and 3?
The factors/ divisors of 2 = 1, 2.
The factors/ divisors of 3 = 1, 3.
Common factors/ divisors = 1
Therefore the highest among common Factors (HCF) = 1 (where 1 is the highest common factor for numbers 2 and 3 since both numbers have only one factor in common.)
Therefore HCF(2, 3) = gcd (2, 3) = 1 is  the required solution of given numbers.

• Those numbers having only one common factor 1 (HCF = 1) are known as Co-Prime Numbers or Relatively Prime. 
• It is not essential that the Co-prime number must be Prime. 
• if we consider few examples like (1, 2), (1, 10), (1, 100), (5, 2), (3, 4), (5, 6), (19, 20) all are Co-Primes since their HCF = 1, although some of them are not Prime.

Various methods to find HCF of Numbers:-

1. By Factorization Method,
2. By Division Method,
3. By Relation between LCM and HCF ( By Formula) 

By  Factorization

The method to find the highest common factor of any given numbers is such as:-

• First of all write down the  Factors or Divisors of individual numbers.
• Find the common factors/ divisors between them. 
• The product of all common factors/ divisors is our HCF or GCD.

Example 2:-  Find HCF of 12 & 18?

Solution:- 

The factors/ divisors of 12 = 1, 2, 3 , 4, 6, 12.

The factors/ divisors of 18 = 1, 2, 3 , 6, 9, 18.
Common Factors = 1, 2, 3, 6.
Therefore the highest among common Factors (HCF) = 6 (where 6 is the highest common factor for numbers 2 and 3 since both numbers have four factors/ divisors in common). 
Therefore HCF(12, 18) = gcd (12, 18) = 6 is the  required solution.

Example 3:-To find HCF of 20, 30 & 50 ( Three numbers)?     

Solution:- 

 Factors of 20 = 1, 2,  4, 5, 10 & 20 ( Since 1x20= 4x5=2x 10).

The factors of 30 = 1, 2,  3, 5, 6, 15, 30.
The factors of 50 = 1, 2,  5, 10, 25 & 50 ( Since 1 x50 = 2 x 25 = 5 x 10).
Common Factors = 1, 2, 5, 10.
Therefore the highest among common Factors (HCF) = 10 (where 10 is the highest common factor for numbers 20,30 and 50 since all numbers have four factors in common).  
Therefore HCF (20, 30, 50) = gcd (20, 30, 50) = 10 is the required solution.

Example 4:-To find HCF of 18,24 & 54?     

Solution :- 

The factors/ divisors of 18 = 1, 2,  3, 6, 9, 18.

The factors / divisors of 24 = 1, 2,  3, 4, 6, 8,12, 24.

The factors / divisors of 54 = 1, 2,  3, 6, 9, 18, 27, 54.

Common factors / divisors = 1, 2, 3, 6.

Therefore the highest among common Factors (HCF) = 6 (where 6 is the highest common factor for numbers 18,24 and 54 since all numbers have four factors/divisors in common.)  

Therefore HCF (18, 24, 54) = gcd (18, 24, 54) = 6 is  the required solution.

 Example 5:-To find GCD / gcd / HCF  of 6 , 16 & 20 ?

Solution :- There are given three numbers 6, 16, and 20. Now we have to write the prime factors ( so do not write 1) of all three numbers individually.

6 =  2 x 3 

16 = 2 x 2 x 2 x 2 x 1

20 = 2 x 2 x 5

The common prime factors divisors of above numbers we get the HCF. Hence, there are ONE pairs of 2 . So the HCF of 6, 16 and 20 will be 2.
HCF (6, 16, 20) = 2   

By Division Method

We have to know about the method of finding the highest common factor using prime factorization or division.  we have to keep in mind that the division method is nothing but dividing the given numbers simultaneously to get the common factors between them. 

ALGORITHMS to solve problems of HCF by are such as.

• First of all write the given numbers horizontally, in a sequence, by separating it with commas ( It is for tradition to separate by commas) order is not important.
• Find the smallest prime number which can divide the all given number. It should exactly divide the all given numbers and we have to write in the left side.
• Now we have to write the quotients carefully.
• We have to repeat the process, unless and until we reach the stage, where there is no further division is possible for all.
•  We will get the common prime factors as the factors in the left-hand side divides all the numbers exactly. 
• The product of these common prime factors is the HCF of the given numbers is our required answers. 

Now Let us solve problems by the above algorithms to find the HCF by division method with the help of these examples "To find HCF of numbers 12, 24, 36".

Example 6:-To find HCF of 12, 24 & 36 ?     

Solution:- 

For solving this problem we have to follow the above mentioned algorithms.

• 

By  Long Division  Method.

Algorithms to find the HCF of given two numbers, For more than three processes can be done similarly taking two at a time in order.

For two numbers:-

• Identify larger and smaller Numbers,
• Divide larger number by smaller number first, such that

Larger Number/Smaller Number

•  Divide the divisor of the above step by the remainder left.

The divisor of the above step /Remainder

•  Again divide the divisor of the above step by the remainder similar to the previous step.

The divisor of the above step /Remainder

• Repeat the process until the remainder is zero.
• The divisor of the last step is the HCF.
• Example 6:-To find HCF of 135 & 225? 
• Solution:-  Since 225 is greater than 135, so we have to divide 225 to 135. 

Therefore HCF (135, 225) = 45 is our required answer.

• For three numbers:-

Repeat the above process taken two at a time in order.

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Least Common Multiple ( LCM )

In arithmetic, mathematics, and number system, the least common multiple ( LCM ) is also known as the smallest common multiple or lowest common multiple ( LCM ) of two integers. For any two integers p and q, usually denoted by LCM(p, q), is the smallest positive integer that is divisible by both p and q.

The LCM is the Lowest Common Denominator (LCD), that can be used before fractions can be added, subtracted, or compared. The LCM of more than two integers  (like for 3 or 4 or 5 any times ) is also well-defined since LCM is the smallest positive integer that is divisible by each of them.

What is the Least Common Multiple(LCM)?
It is a method to find the smallest common multiple between numbers ( any two or more). Generally according to definitions the common multiple is a number which is a multiple of two or more numbers therefore LCM is used to determine the least common factor or multiple  ( Generally called as Tables) of an integer. Since the division of integers by zero is undefined or not possible, this definition has meaning only if a and b are both different from zero  (so not zero ). 
Example 1:-
If we consider an example,  To find L.C.M of 2 and 3?
Now multiples of 2 = 0, 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22 , 24...
Again Multiples of 3 can be  written as  = 0, 3, 6, 9,  12, 15, 18, 21, 24, 27,30, 33. 36, 39.42. 45, 48. ...
Common Multiples = 0, 6, 9, 18, 24, ....
Therefore least among common Factors ( LCM) = 6 ( where 6 is the smallest common multiple for numbers 2 and 3.)

Algorithm to find LCM of Numbers:-

1. By Finding the Multiples,
2. By Prime Factorization
3. By Multiple Tree Diagram.

4. By Relation Between LCM and HCF ( By Formula ).                                                                                                                                    1 By Finding the Multiples

   The method to find the least common multiple of any given numbers is first to write down the multiples (Tables) of individual numbers and then find the first common multiple between them. 
   Example 2:-  To find LCM of 3 & 4 ?                                                                                                                  Solution:-     Multiples of 3 = 3, 6, 9,  12, 15, 18, 21, 24 ...                                                                  Multiples of 4 = 4, 8, 12, 16, 20, 24, 28, 32,   ….                                                                                                 Common Multiples = 12, 24, ....
   Therefore least among common Factors ( LCM) = 12 ( where 12 is the smallest common multiple for numbers 3 and 4. )                                                                                                                                              Example 3:-To find LCM of 3 4 & 6 ( Three numbers)?

Solution:- 

Multiples of 3 = 3, 6, 9,  12, 15, 18, 21, 24 , 27,30, 33...                                                                                                   Multiples of 4 = 4, 8, 12, 16, 20, 24, 28, 32, 36,40, 44, 48,….                                                                                                        Multiples of 6 =   6, 12, 18, 24, 30, 36, 42, 48, 54...                                                                                                                Common Multiples = 12, 24,36 ....

Therefore least among common Factors ( LCM) = 12 ( where 12 is the smallest common multiple for numbers 3, 4 and 6.)   

2 By Prime Factorization

Most Popular method to find the LCM of the given numbers is by prime factorization. 

 Example 4:-To find LCM of 6 16 & 20 ( Three numbers)?Solution :- There are given three numbers 6, 16, and 20. Now we have to write the prime factors of all three numbers individually.

6 =  2 x 3 ==  2 x 3 x 1 ( Also written as )

16 = 2 x 2 x 2 x 2 = 2 x 2 x 2 x 2 x 1 ( Also written as ) 

20 = 2 x 2 x 5 = 2 x 2 x 5 x 1 ( Also written as )

On pairing the common prime factors of above numbers we get the LCM. Hence, there are four pairs of 2 and one pair of 3 and 5 each. So the LCM of 6, 16 and 20will be;

LCM (6, 16, 20) = 2 x 2 x 2 x 2 x 3 x 5  = 240                                                                                                                                      3 By Multiple Tree Diagram.

Example 5:-  To find LCM of 6 & 10?    

The multiple trees can be formed by using the prime factorization method this is similar to the above method. 

                 

Therefor LCM of 6 and 10 = 1 x 2 x 3 x 5 = 30.

this method is known as Tree Diagram.

Use's of LCM:- 
LCM can be used before the addition, subtraction, or comparison of fractions can take place, Therefore it is basic fundamental tools for using mathematical operations in Fractions.
The LCM of more than two integers also happens to be well-defined: it is the smallest positive integer whose division can take place by each of them.

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NTSE ( National Talent Search Examination)

   

National Talent Search Examination

To identify and nurture talented students across the country along with Indian students studying abroad The "National Talent Search Examination: popularly known as 'NTSE' is a  unique National Level scholarship program in India for students studying in 10 class organized By the Government of India by ( NCERT as a Nodal agency) 
It recognizes the young talents of India and honors them also helps talented students by providing continuous financial assistance in the form of a monthly scholarship for the entire Academic Career.
NTSE  Examination conducted by the National Council of Educational Research and Training (NCERT) for the second Phase and first phase by Boards Or respective nodal agencies of states.

This examination is divided into two papers  MAT ( Mental Ability Test ) and SAT (Scholastic Aptitude Test).

For the preparation of  NTSE following books may be consulted/suggested (as per my opinions ) according to subject wise are as follows:-

 Paper - I     
MAT ( Mental Ability Test ) ( IMPORTANT PART )

1. Any book according to your choice few topics may be followed from  R. S. Agarwal
2. Mental Ability for NTSE stage 1 & 2  by Disha Publication is also concise and useful.

Paper-II Scholastic Aptitude  Test (SAT) 

1. Social science:-
1. NCERT book of 9 and 10 with all parts.
2. Some latest general knowledge questions ( 1 or 2 questions may or may not be asked).
3. Stress on maps and facts and figures are given in the books.
2. Science:-
1. NCERT books of 9 and 10.
2. NCERT Exemplars.
3. Some reference books according to your choices like  H. C. Verma for Physics ( 9 and 10 Both classes ) and Lakhmir Sing & Kaur for Biology and Chemistry.
3. Mathematics: -
1. NCERT books of 9 and 10.
2. NCERT Exemplars.
3. Solve a few advanced problems in geometry and number system by any book like R. D. SHARMA / R, S, AGARWAL ETC.

• Solve last 5 Year solved papers of BOTH stages along with few modal papers from any standard reference book  ( Arihant publication or TMH or etc ) for time management.                                                                                                                                                              Process and pattern of examination along with useful information are such as:-

1. How to apply:-The students studying in Class X in the country ought to be on the lookout for any advertisement in the newspapers or circulars in the school by the respective Government of their State/UT for the above said examination and act as per the requirement given in the State advertisement/circular in the beginning of schools. Any other information/query about the details of the state level examination maybe had from Liaison Officers of the State/UT agencies may be obtained generally their state birds are nodal agencies. The filled-in application by the student be submitted to the State Liaison Officer {may be changed from time to time (http://www.ncert.nic.in/programmes/talent_exam/pdf_files/LO_list_2019-20.pdf)}  duly signed by the Principal of the school in which studying 10 classes before the last/due date as advertised/circulated by their respective  State/UT.
2. For Indian Students Studying Abroad:- Studying in class X  can directly apply to NCERT for second stage NTS examination under conditions prescribed by NTSE in advertisement brochure up to 31DECEMBER ( May be changed see on NCERT web regularly).
3. The medium of Examination:-The medium of the test shall be as announced by their respective  State/UT. 
4. Fee :- State may impose any fee for examination and/or for application form while for the second stage no fees. 
5. Examination:- State level examination also has two parts like to second stage (but you may check before appearing in examinations ): Part-I Mental Ability Test (MAT) and Part II Scholastic Aptitude Test (SAT) for nominating the required number of candidates ( already fixed foe each state and union territories) for the second level test to be conducted by the NCERT.   
6. Written Examination:-The written examination shall consist of two paper; Paper-I Mental Ability Test (MAT) and Paper-II Scholastic Aptitude Test (SAT). Both the  papers (tests) will be administered on the same date
7. Mental Ability Test:-There shall be 100 multiple-choice type items, with four alternatives. Each item will carry an equal ( one) mark. Candidates are required to answer the items on a separate OMR ( Optical Mark Recognition) sheet as per instructions given in the test booklet and on the OMR sheet.
8. Scholastic Aptitude Test:-The scholastic Aptitude Test will consist of 100 multiple-choice problems/questions of one mark each. Each problem/question shall have four alternatives/options, out of which only one will be the correct answer. There shall be 40 problems/questions s from Science, 40 from Social Science and 20 from Mathematics, Candidates are required to answer the items on a separate OMR sheet given at the examination center. For the convenience of students, they are allowed to take away question booklets of both the tests (Mental Ability Test and Scholastic Aptitude Test) along with a copy of the OMR sheet after the examination.
9. Medium:- The tests will be available in the following languages: Asamiya, Bangla, English, Gujarati, Hindi, Kannada, Marathi, Malayalam, Odia, Punjabi, Tamil, Telugu and Urdu. 
10. Venue, Date, and Time of the Test:-The admission letter shall have all the information about the venue, the roll number, the date and time of the test. Candidates can download their E-Admit cards themselves before 21 days 2 or 3 weeks of examination.
11. FOR the FIRST STAGE:- The state-level  stage first screening examination is conducted in all State/UT’s on the first Sunday of November  (You may check )except in Nagaland, Andaman and Nicobar Island, Meghalaya and Mizoram where it will be generally  conducted on first Saturday of November every year (You may check )until and unless some special circumstances occur
12. Syllabus:-There is no prescribed syllabus for the National Talent Search Examination (NTSE). However, the standard of problems/questions shall be conforming to the level of Classes IX and X. Information about test and problems may be obtained from (separate booklet called ‘Learn about the Test’ containing sample items for both the Tests- MAT and SAT is available  on the NCERT website ) along with in some standard Guides.
13. Centre of Examination: Generally the candidate belonging to a particular state shall be allotted the center in their state where studying 10 class. Normally there are one/two/ three / four centers ( According to the number of students and region of sate )in each state for the second stage National Level Examination. The conduct of examination at the centers shall be done under the supervision of the Centre Superintendent (CS) appointed by the NCERT. 
14. Scholarship  At present National Talent Search Examination (NTSE)  Scholarship Amount are such as:-

    Stage	Scholarship Amount
    Higher Secondary level (for class 11 & 12)	Rs.1250/-
    Graduate and Post Graduate (in all streams like B, Tec / B. E./ M. Tech/ MBBS/MS/ B. Sc./ B. A./B. Com./ BBA/ MBA/ etc.)	Rs.2000/-
    For a Ph. D. degree (four years) in all disciplines.	As per UGC norms for JRF Follows.

15. Selection Criteria:-• Only candidates qualifying in both the papers separately will be considered for merit. 

• Selection of the awardees will be made on the basis of total Marks scored in MAT & SAT based on merit according to their respective categories.
•  There will be No Negative Mark
•  Qualifying marks 32% SC, ST and PWD and 40% for Others (General& OBC)    ___________________________________________________________________________