*H.C.F AND L.C.M**Facts And Formulae:*

1.Highest Common Factor:(H.C.F) or Greatest Common Meaure(G.C.M) :

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 :

1.Factorization method: Express each one

of the given numbers as the product of prime

factors.

The product of least powers of common prime factors gives HCF.

Example : Find HCF of 26 * 32*5*74 , 22 *35*52 * 76 , 2*52 *72

Sol: The prime numbers given common numbers are 2,5,7

Therefore HCF is 22 * 5 *72 .

**2.Division Method**: Divide the larger number by

smaller one. Now divide the divisor by

remainder. Repeat the process

of dividing preceding number last obtained till

zero is obtained as

number. The last divisor is HCF.

Example: Find HCF of 513, 1134, 1215

Sol:

1134) 1215(1

1134

----------

81)1134(14

81

-----------

324

324

-----------

0

-----------

HCF of this two numbers is 81.

81)513(6

486

--------

27)81(3

81

-----

0

---

HCF of 81 and 513 is 27.

**3.Least common multiple[LCM]**: The least number which is

divisible by each one of given

numbers is LCM.

There are two methods for this:

**1.Factorization method**:

Resolve each one into product of prime factors.

Then LCM is product of highest powers

of all factors.

2.Common division method.

Problems:

1.The HCF of 2 numbers is 11 and LCM is 693.If one

of numbers is 77.find other.

Sol: Other number = 11 * 693/77=99.

2.Find largest number of 4 digits divisible by 12,15,18,27

Sol: The largest number is 9999.

LCM of 12,15,18,27 is 540.

on dividing 9999 by 540 we get 279 as remainder.

Therefore

number =9999 â€“ 279 =9720.

3.Find least number which when divided by 20,25,35,40

leaves remainders 14,19,29,34.

Sol:

20â€“14=6

25-19=6

35-29=6

40-34=6

Therefore number =LCM of (20,25,35,40) - 6=1394

4.252 can be expressed as prime as :

2 252

2 126

3 63

3 21

7

prime factor is 2 *2 * 3 * 3 *7

5.1095/1168 when expressed in simple form is

1095)1168(1

1095

------

73)1095(15

73

---------

365

365

---------

0

----------

So, HCF is 73

Therefore

1095/1168 = 1095/73/1168/73= 15/16

6.GCD of 1.08,0.36,0.9 is

Sol:

HCF of 108,36,90

36)90(2

72

----

18)36(2

36

----

0

----

HCF is 18.

HCF of 18 and 108 is 18

18)108(6

108

-------

0

--------

Therefore HCF =0.18

7.Three numbers are in ratio 1:2:3 and HCF is 12.Find numbers.

Sol:

Let the numbers be x.

Three numbers are x,2x,3x

Therefore

HCF is

2x)3x(1

2x

-----

x)2x(2

2x

--------

0

-------------

HCF is x so, x is 12

Therefore numbers are 12,24,36.

8.The sum of two numbers is 216 and HCF is 27.

Sol:

Let numbers are

27a + 27 b =216

a + b =216/27=8

Co-primes of 8 are (1,7) and (3,5)

numbers=(27 * 1 ), (27 * 7)

=27,89

9.LCM of two numbers is 48..The numbers are in ratio 2:3.

The sum of numbers is

Sol:

Let the number be x.

Numbers are 2x,3x

LCM of 2x,3x is 6x

Therefore

6x=48

x=8.

Numbers are 16 and 24

Sum=16 +24=40.

10.HCF and LCM of two numbers are 84 and 21.If ratio of

two numbers is 1:4.Then largest of two numbers is

Sol:

Let the numbers be x,4x

Then x * 4x = 84 * 21

x2 =84 * 21 /4

x = 21

Largest number is 4 * 21.

11.HCF of two numbers is 23,and other factors of LCM are

13,14.Largest number is

Sol:

23 * 14 is Largest number.

12.The maximum number of students among them 1001 pens

and 910 pencils can be distributed in such a way that

each student gets same number of pens and pencils is ?

Sol:

HCF of 1001 and 910

910)1001(1

910

------------

91)910(10

910

--------

0

---------

Therefore HCF=91

13.The least number which should be added to 2497 so that sum is

divisible by 5,6,4,3 ?

Sol: LCM of 5,6,4,3 is 60.

On dividing 2497 by 60 we get 37 as remainder.

Therefore number to added is 60 â€“ 37 =23.

Answer is 23.

14.The least number which is a perfect square and is divisible by

each of numbers 16,20,24 is ?

Sol: LCM of 16,20,24 is 240.

2 * 2*2*2*3*5=240

To make it a perfect square multiply by 3 * 5

Therefore 240 * 3 * 5=3600

Answer is 3600.

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