Highest Common Factor (H.C.F.) or Greatest Common Measure (G.C.M.) or Greatest Common Divisor (G.C.D.):

Factors and Multiples:

If number 'x' divided another number 'y' exactly, we say that 'x' is a factor of 'y'.

In this case, 'y' is called a multiple of 'x'.

H.C.F. - The H.C.F. of two or more than two numbers is the greatest number that divides each of them exactly.

Highest Common Factor (HCF)

The greatest number that divides each of the given numbers exactly is called HCF.

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

(1) Factorization Method: Express the each one of the given numbers as the product of prime factors. The product of least powers of common prime factors gives H.C.F.

(2) 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.

Finding the H.C.F. of more than two numbers:  Suppose we have to find the H.C.F. of three numbers, then, H.C.F. of [(H.C.F. of any two) and (the third number)] gives the                                                                                             H.C.F. of three given number.

Similarly, the H.C.F. of more than three numbers may be obtained.

Lowest Common Multiple (LCM)

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

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

(1) Factorization Method: Resolve each one of the given numbers into a product of prime factors. Then, L.C.M. is the product of highest powers of all the factors.

(2) Division Method (short-cut):  Arrange the given numbers in a rwo 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.

(1) Product of two numbers = Product of their H.C.F. and L.C.M.

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

(3) H.C.F. and L.C.M. of Fractions:

(4) H.C.F. and L.C.M. of Decimal Fractions:

In a given numbers, make the same number of decimal places by annexing zeros in some numbers, if necessary. Considering these numbers without 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.

(5) 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.

METHODS OF LCM

(1) Factorization method        and       (2) Division method

1. Factorization method: -

• Express each one of the given numbers as the product of prime factors.
• The product of the highest powers of all the  factors gives LCM.
  Ex: Find the LCM of (48,108,140)

       ⇒Prime factors of 108 = 2²×3³
        ⇒ Prime factors of 48 = 2⁴×3
         ⇒ Prime factors of 140 = 2²×5×7
         ⇒LCM = 2⁴×3³×5×7 =15120

2. Division method

        Ex: Find the LCM of (120, 144, 160, 180)

        

⇒LCM = 2⁴ ⅹ 3² ⅹ 5 ⅹ 2   = 1440
TYPES OF QUESTIONS ON LCM

(1) LCM of integers   (2) LCM of decimals   (3) LCM of Fractions         (4) LCM of algebraic expression

LCM of integers: -      We have to do the same normal calculation as above.
LCM of decimals: -     We have to convert decimal into integers by making the number of digits after decimal point is common, now find LCM normally and add decimal point

Ex- LCM of (0.16, 0.48)
⇒Now the decimal is converted into integers and we get is (16, 48)
⇒Now do LCM and we get the answer = 2⁴ × 3 = 48

⇒Now the number 48 is converted into decimal 0.48

LCM of fractions: -    

The Remainder Based Problem In LCM

Case-1. When the remainder is same, i.e. “R”

 TRICK ⇢ LCM [n₁, n₂, n_3, ……..] + R

Ex-     Find the LCM of (8,9,12,15) that leaves same remainder 1 in each the case.
Solution:          ⇒LCM of [8,9,12,15]
⇒2³×3²×5         ⇒360   ⇒à [360] + 1
⇒Now we get the answer as 361 

Case-2. When the remainder is different in each case

 TRICK 1⇢ n₁- R₁ = n₂-R₂ = n₃-R₃ = R
TRICK 2⇢ LCM (n1, n2, n3) – R

Ex-     Find the LCM of (18, 27, 36) that leaves different remainders (5, 14, 23)
Solution: 

⇒18-5=13 ,      ⇒27-14=13      and ⇒36-23=13

⇒So now R=13
⇒LCM (18, 27, 36) = 3³×2² =108
⇒Now we subtract the number 108 with R⇢ [108-13=95]
⇒So the answer is 95.
Exercise with Solution

1. Find the smallest number which when diminished by 7 is divisible by 12,16,18,21 and 28?

Solution: -

Required no = LCM (12,16,18,21 and 28) + 7

 = 1008 + 7= 1015 Ans.

2.  Find the least number which divided by 48, 60, 72, 108 and 140 leaves 38, 50, 62, 98 and 130 as remainders respectively.

Solution: - 

48-38=10, 60-50=10,        72-62=10,

108-98=10,            140-130=10

Required no = LCM (48,60,72,108, and 140) - 10

 = 15120 - 10 = 15110 Ans.

3. The largest four-digit number which when divided by 4,7 or 13 leaves a remainder of 3 in each case? 

Solution

The largest 4 digits no = 9999

LCM of 4,7,13 = 364
Now divide 9999 by 364 and we get the remainder as 171
4 digit divisible by 4,7,13=(9999-171)=9828
Now add the remainder 3 given in question (9828+3)
⇒ And answer is 9831.

4. Six bells commence tolling together and toll at intervals of 2,4,6,8,10 and 12 seconds respectively. In 30 minutes how many times do they toll together? 

Solution

⇒ LCM of 2,4,6,8,10 and 12 = 120
⇒ So the bell will toll together every 120sec = 2min
⇒ In 30 minutes they will toll together (30/2) +1
⇒ And the answer is 16 times.

6. Traffic lights at three different road crossing change after every 48 sec, 72sec, and 108sec respectively. If they all change simultaneously at 8:20:00 hours then they will again change simultaneously at what time?
Solution

⇒ LCM of 48,72,108 = 432 sec
⇒ Now 432 sec =7 min 12 sec
⇒ Now add the time (8hr 20min 0sec) + (7min 12 sec)
⇒ so signal will again change simultaneously at 8:27:12.

Highest Common Factor (HCF) GDF

The HCF of two or more than two numbers are the greatest number that divides each of them exactly.

Methods to find HCF

1. Factors Method  
2. Prime factorization     
3. Long division   
4. Difference method
5. Euclid method

HCF by Factors method

Express each one of the given numbers as factors. The highest common factor gives HCF

Ques-   Find the HCF of (36,48)

Factor of 36= 1, 2, 3, 4, 6, 9, 12, 18 and 36

Factor of 48= 1, 2, 3, 4, 6, 8, 12, 16, 24 and 48

Hence HCF (36 & 48) = Highest common factor

                                    = 12 Ans.

HCF by Prime factorization method:

Express each one of the given numbers as the product of prime factors. The product of least powers of common prime factor gives HCF
Example

Find the HCF of (36,48)
⇒ 36=2² х 3²
⇒ 48=2⁴ х 3
⇒ So 2² х 3 is the common least factor and the answer is 12

HCF by Long division method

Example

Find the HCF of (16, 24)

So the answer is 8.
Difference method

• Find the difference of any two of given numbers.
• If difference divides the no then it will be ans.
• If difference don’t divide the no then write the factor of difference.
• HCF= greatest factor that divides the no

Ques-   Find the HCF of (36,48)

            Difference= 12

            HCF= 12 because it divides the nos.

Ques-   Find the HCF of (15, 36,48)

            Difference of 15 & 36 = 21( it don’t divides)

            Factor of 21 = 1, 3, 7, 21

            HCF= 3 because it divides the nos 15, 36, 48.

TYPES OF QUESTIONS ON HCF

(1) HCF of integers  (2) HCF of decimals  (3) HCF of Fractions   (4) HCF of algebraic expression
HCF of integers: -       We have to do the same normal calculation as above
HCF of decimals: -      We have to convert decimal into integers by making the number of digits after decimal point is common

Example

(0.54,0.06,1.08)

⇨Let us multiply each number by 100

⇨We get 54, 6 and 108

⇨Now the HCF of 54, 6 and 108 = 6

⇨Now divide 6 by 100

⇨We get 0.06

⇨ So the answer is 0.06

HCF of fractions: -     

The Remainder Based Problem In HCF

Case 1: Same remainder in each case

 Trick ⇢           HCF [(n₂-n₁) (n₃-n₂) (n₃-n₁)] 

Example

Find the greatest number that can divide (62,132,237) that leaves same remainder in each case
Solution: 

HCF of [(132-62) (237-132) (237-62)]
HCF of [70,105,175]
Now we take HCF of these numbers and we get the answer as 35.

Case 2: Same remainder “R” 

 Shortcut⇨ HCF [(n₁-R) (n₂-R) (n₃-R)] 

Example

Find the HCF of (201,333,339) that leaves same remainder 3 in each the case
Solution: 

HCF of [(201-3) (333-3) (339-3)]
HCF of [(198) (330) (396)]
Now we take HCF of these numbers and we get the answer as 66

Case 3: Different remainder in each case

 Shortcut⇨ HCF [(n₁-R₁) (n₂-R₂) (n₃-R₃)] 

Example

Find the HCF of (42,49,56) that leaves different remainders (0,1,2)
Solution:

HCF of [(42-0) (49-1) (56-2)]
HCF of [42, 48, 54]

Now we take HCF of these numbers and we get the answer as 6.

Ratio Based Problem In HCF

Example-  A and B are in the ratio 2:3. If their HCF is 13, then find the values of A and B?
Solution: 

Now we apply the formula 2:3 = Value/13

Value =13×2, 13×3
Value of A and B =26, 39

Ques 1:  Two numbers are in the ratio 4:7 and their HCF is 15. Find the difference between the two numbers?

Solution:  We have to use ratio formula and we will get two numbers 60,105 and its difference is 45

Ques 4: The sum of two numbers is 216 and their HCF is 27. How many pairs of such numbers are there?

Solution: 

In this two numbers x and y and hcf is 27 

27x+27y=216 Or 27(x + y) = 216 

X + y=216/27 Or x + y=8

Now we check prime multiplication and it is

7+1=8 ⇾1st pair coprime 

5+3=8⇾2nd pair 

4+4=8⇾ not right pair