HCF & LCM

H.C.F and L.C.M

Factors and Multiples:

If a number `a' divides another number `b' exactly, then we say that `a' is a factor of `b' and that `b' is a multiple of `a'.

eg. 2 is a factor of 12 and therefore 12 is a multiple of 2.

Least Common Multiple (L.C.M.)

L.C.M. is the least non-zero number in common multiples of two or more numbers.

Multiple of 6 =  6, 12, 18, 24, 30, ....

Multiple of 8 =  8, 16, 24, 32, 40, ........

Common Multiple of 6 and 8 = 24, 48

Least Common Multiple = 24

Factorisation Method:

Q-Find the L.C.M. of 12, 27 and 40 ?

Factors of 12 = 2 * 2*3

Factors of 27 = 3 *3*3

Factors of 40 = 2*2*2*5

L.C.M = 2*2*2*3*3*3*5 = 1080

SHORT CUT METHOD

(Division Method)

Q-Find the L.C.M. of 12, 27, 40 ?

2      12,27,40

2       6,27,20

3       3,27,10

         1,9,10

 L.C.M = 2*2*3*9*10 = 1080

HIGHEST COMMON FACTOR (H.C.F)

The highest common factor of two or more numbers is the greatest number which divides each of them exactly.

Q- Find the H.C.F. of  24 , 56 ?

Factors of 24 = 1, 2, 3, 4, 6, 8, 12, 24

Factors of 56 = 1, 2, 4, 7, 8, 14, 28, 56

Common factors of 24 and 56 are 1, 2, 4, 8

H.C.F. of 24 and 56 = 8

Factorisation Method:

H.C.F. can be found by resolving the given numbers into prime factors and then taking the product of least powers of all common factors, that occur in these numbers.

Q- Find H.C.F. of 48, 108, 140 ?

Factors of 48 = 2*2*2*2*3                                                                                                                                                       Factors of 108 = 2*2*3*3*3

Factors of 140 = 2*2*5*7

H.C.F. = 2*2 = 4

HCF OF GIVEN FRACTIONS:

H.C.F of given fractions = HCF of numerator/LCM of denominator

Example 1:  Find the H.C.F of 4/9 ,16/15 , 12/21 ?

Solution:

According to the formula, H.C.F of numerators 4, 16, 12

4 = 2x2

16=2x2x2x2

12=2x2x 3

So, the H.C.F of numerators = 4

L.C.M of denominators 9, 15, 21

9 = 3x3

15= 3x5

21= 3x7

So, the L.C.M of denominators = 3*3 *5*7 = 315

Therefore, H.C.F of given fractions= 4/315

L.C.M OF GIVEN FRACTIONS:

L.C.M of given fractions =  LCM of numerator/HCF of denominator

Example 2:   Find the L.C.M of the fractions 2/3,  7/18 , 11 /12?

Solution:

According to the formula, L.C.M of numerators 2, 7, 11.

 As the numerators are all prime numbers, L.C.M of numerators = 2*7*11 = 154

H.C.F of denominators 3, 18, 12

3=3

18=3*3*2

12=2*2*3

So, H.C.F of denominators 3, 18, 12 = 3

Therefore, L.C.M of given fractions = 154/3 

The product of two given numbers is equal to the product of their H.C.F. and L.C.M.

Example 3:   The L.C.M. of two number is 2310. Their H.C.F. is 30. If one number is 210, the other number is

Solution:

The other number =  (L.C.M * H.C.F ) / given number = (2310 * 30) / 210 = 330

Type: The greatest number which divides x, y and z to leave the remainder R is H.C.F of (x – R), (y – R) and (z – R).

Example 4: Find the greatest number divides 24, 60, and 84 leaves the remainder 3?

Solution:

By definition the greatest number which divides the given numbers and leaves a remainder of 0, only if it is the H.C.F of given numbers

x = 24, y = 60, z = 84 and R = 3

H.C.F of (x - R), (y - R), (z - R) = H.C.F of 21, 57, 81

21= 3*7

57 =3*19

81 = 3*3*3*3

So, H.C.F of 21, 57, 81 =3.

 Therefore the greatest number is 3 which divide 24, 60 and 84 to leave a remainder 3

Type:  The greatest number which divide x, y, z to leave remainders a, b, c is H.C.F of (x - a), (y - b) and (z - c).

Example 5: Find the greatest number which divides 18, 26 and 54 to leave remainders 3, 1, 4?

Solution:

Here, x = 18, y = 26, z = 54, a = 2, b = 1, c = 4

H.C.F of (x - a), (y - b), (z - c) = H.C.F of 15, 25, 50

 15= 5*3

 25= 5*5

 50= 2*5 *5

So, H.C.F of 15, 25, 50 = 5, we can cross check by dividing x, y, z and obtain the same remainders as mentioned in question.

Type: The smallest number which when divided by x, y and z leaves remainder of a, b, c (x – a), (y - b), (z - c) are multiples of K

Required number = (L.C.M of x, y and z) – K

Example 6: Find the Smallest number which divides 2, 5, 7 to leave remainders 0, 3, 5?

Solution:

Here, x = 2, y = 5, z = 7, a = 0, b = 3, c = 5

 (x - a), (y - b), (z - c) = 2, 2, 2

Therefore, K = 2

L.C.M of x, y, and z = 2 * 5 * 7 = 70

 Required number = (L.C.M of x, y and z) – K = 70 -2 =68

Example 7: What is the least number which when divided by 8,9,12 and 15 leaves the same remainder 1 in each case ?

Solution:

Required number = ( l.c.m of 8,9,12,15)+1 = 361

Example 8:The traffic lights at three different road crossings change after every 48 sec, 72 sec and 108 sec respectively. If they all change simultaneously at 8 : 20 : 00 hours, then they will again change simultaneously at?

Solution:

Interval of change = (l.c.m of 48,72,108) = 432 sec. The lights will change simultaneously after every 432 seconds . Next simultaneous change will take place at 8:27:12 hrs.