What is Facts and Formulae of L.C.M & H.C.F?

 H.C.F. AND L.C.M. OF NUMBERS: IMPORTANT FACTS AND FORMULAINMPORTANT FACTS AND FORMULA

I. Factors and Multiples : If a number x divides another number y exactly, then, we say that x

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

II. Highest Common Factor (H.C.F.) or Greatest Common Measure (G.C.M.) or

Greatest Common Divisor (G.C.D.): The H.C.F. of two or more than two numbers is

the greatest number that divides each of them exactly.

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

1. Factorization Method:- In this method, we express 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: In this method, we follow method and steps.

Suppose that we have to find the H.C.F. of two given numbers. We shall divide the larger number by the smaller one in first step. In second step, we shall divide the divisor by the remainder. Again repeat the process of dividing the preceding number by the remainder last obtained till zero is obtained as remainder. The last divisor is the required H.C.F.

FINDING THE H.C.F. OF MORE THAN TWO NUMBERS:- Suppose that we have to find the

H.C.F. of three numbers. The question rises in your mind is will In this case, we shall get H.C.F of any two numbers first then get H.C.F of first two numbers and third number. Similarly, the H.C.F. of more than three numbers can be obtained.

          

III. Least Common Multiple (L.C.M.) : The least number which is exactly divisible by

each one of the given numbers is called their L.C.M. There are two method of finding L.C.M.

1. Factorization Method of Finding L.C.M.:- In this method, we resolve each one of the given

numbers into a product of prime factors. Then, get L.C.M. is the product of highest

powers of all the factors.

2. Common Division Method {Short-cut Method) of Finding L.C.M.:

In this method we arrange the given numbers in a row in any order. Divide all these numbers by a certain number which divides exactly at least two of the given numbers and carry forward those 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. e.g. find L.C.M of given number 3, 5, 6, 9, 12, 14, 15, 16, 18, 20.

Solution: When we start to find out L.C.M of given numbers, we select least number which can divides maximum numbers, and then the bigger number. Here, we will starts with 2 first and then 3.

 2|3, 5, 6, 9, 12, 14, 15, 16, 18, 20

 2|3, 5, 3, 9, 6, 7, 15, 8, 9, 10       

 3|3, 5, 3, 9, 3, 7, 15, 4, 9, 5

 3|1, 5, 1, 3, 1, 7, 5, 4, 3, 5,

 3|1, 5, 1, 1, 1, 7, 5, 4, 1, 5,

 5|1, 1, 1, 1, 1, 7, 1, 4, 1, 1

Since, now, there is no number left more than one. So that we shall all numbers and get L.C.M.

2 x 2 x 3 x 3 x 3 x 5 x 7 x 4 x 1 = 4 x27 x 35 x4

16 x 27 x 35 = 15120 required number or L.C.M

IV. Product of two numbers =Product of their H.C.F. and L.C.M.

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

VI. H.C.F. and L.C.M. of Fractions:

1.H C F= H.C.F. of Numerators 2.L C M = L.C.M of Numerators__

L.C.M. of Denominators H.C.F. of Denominators

VII. 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 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.

VIII. 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.

SOLVED EXAMPLES

Ex. 1. Find the H.C.F. of 23 X 32 X 5 X 74, 22 X 35 X 52 X 73, 23 X 53 X 72

Sol. The prime numbers common to given numbers are 2,5 and 7.

H.C.F. = 22 x 5 x72 = 980.

Ex. 2. Find the H.C.F. of 108, 288 and 360.

Sol. 108 = 22 x 33, 288 = 25 x 32 and 360 = 23 x 5 x 32.

H.C.F. = 22 x 32 = 36.

Ex. 3. Find the H.C.F. of 513, 1134 and 1215.

Sol.   1134 ) 1215 ( 11134

              81 ) 1134 ( 14

                        81

                        324

                        324

                    x

   H.C.F. of 1134 and 1215 is 81.

So, Required H.C.F. = H.C.F. of 513 and 81.

______

81 ) 513 ( 6

   __486____

   27) 81 ( 3

         81

           0

H.C.F. of given numbers = 27.

Ex. 4. Reduce 391 to lowest terms .667 to lowest terms.

Sol. H.C.F. of 391 and 667 is 23.

On dividing the numerator and denominator by 23, we get :

391 = 391 23 = 17

667 66723 29

Ex.5.Find the L.C.M. of 22 x 33 x 5 x 72 , 23 x 32 x 52 x 74 , 2 x 3 x 53 x 7 x 11.

Sol. L.C.M. = Product of highest powers of 2, 3, 5, 7

and        11 = 23 x 33 x 53 x 74 x 11

Ex.6. Find the L.C.M. of 72, 108 and 2100.

Sol. 72 = 23 x 32,

      108 = 33 x 22, 2100

              = 22 x 52 x 3 x 7.

L.C.M. = 23 x 33 x 52 x 7 = 37800.

Ex.7.Find the L.C.M. of 16, 24, 36 and 54.

Sol.

2| 16 - 24 - 36 - 54

2| 8 - 12 - 18 - 27

2 |4 - 6 - 9 - 27

3 |2 - 3 - 9 - 27

3 |2 - 1 - 3 - 9

2 |- 1 - 1 - 3

L.C.M. = 2 x 2 x 2 x 3 x 3 x 2 x 3 = 432.

Ex. 8. Find the H.C.F. and L.C.M. of 2 , 8 , 16 and 10.3 9 81 27

Sol. H.C.F. of given fractions = H.C.F. of 2,8,16,10 = 2_

L.C.M. of 3,9,81,27 81

L.C.M of given fractions = L.C.M. of 2,8,16,10 = 80_

H.C.F. of 3,9,81,27 3

Ex. 9. Find the H.C.F. and L.C.M. of 0.63, 1.05 and 2.1.

Sol. Making the same number of decimal places, the given numbers are 0.63, 1.05 and

2.10.

Without decimal places, these numbers are 63, 105 and 210.

Now, H.C.F. of 63, 105 and 210 is 21.

H.C.F. of 0.63, 1.05 and 2.1 is 0.21.

L.C.M. of 63, 105 and 210 is 630.

L.C.M. of 0.63, 1.05 and 2.1 is 6.30.

Ex. 10. Two numbers are in the ratio of 15:11. If their H.C.F. is 13, find the

numbers.

Sol. Let the required numbers be 15.x and llx.

Then, their H.C.F. is x. So, x = 13.

The numbers are (15 x 13 and 11 x 13) i.e., 195 and 143.

Ex. 11. TheH.C.F. of two numbers is 11 and their L.C.M. is 693. If one of the

numbers is 77,find the other.

Sol. Other number = 11 X 693 = 99 x 77

Ex. 12. Find the greatest possible length which can be used to measure exactly the

lengths 4 m 95 cm, 9 m and 16 m 65 cm.

Sol. Required length = H.C.F. of 495 cm, 900 cm and 1665 cm.

                            

                              495 = 32 x 5 x 11,

                              900 = 22 x 32 x 52

                            1665 = 32 x 5 x 37.

                        H.C.F. = 32 x 5 = 45.

Hence, required length = 45 cm.

Ex. 13. Find the greatest number which on dividing 1657 and 2037 leaves

remainders 6 and 5 respectively.

Sol. Required number = H.C.F. of (1657 - 6) and (2037 - 5) = H.C.F. of 1651 and 2032

_______

1651 ) 2032 ( 1 1651

_______

               381 ) 1651 ( 4

1524_________

127 ) 381 ( 3

381

0

Required number = 127.

Ex. 14. Find the largest number which divides 62, 132 and 237 to leave the same

remainder in each case.

Sol . Required number = H.C.F. of (132 - 62), (237 - 132) and (237 - 62)

= H.C.F. of 70, 105 and 175 = 35.

Ex.15.Find the least number exactly divisible by 12,15,20,27.

Sol.

3| 12 - 15 - 20 - 27

4 |4 - 5 - 20 - 9

5 |1 - 5 - 5 - 9

1 | 1 - 1 - 9

Ex.16.Find the least number which when divided by 6,7,8,9, and 12 leave the same

remainder 1 each case

Sol. Required number = (L.C.M OF 6,7,8,9,12) + 1

3| 6 - 7 - 8 - 9 - 12

4 |2 - 7 - 8 - 3 - 4

5 |1 - 7 - 4 - 3 - 2

1 | 7 - 2 - 3 – 1

L.C.M = 3 X 2 X 2 X 7 X 2 X 3 = 504.

Hence required number = (504 +1) = 505.

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