Solved Examples

1. LCM of 27,314and53is${\displaystyle \frac{2}{7},{\displaystyle \frac{3}{14}\mathrm{a}\mathrm{n}\mathrm{d}{\displaystyle \frac{5}{3}\mathrm{i}\mathrm{s}}}}$

a. 45

b. 35

c. 30

d. 25

Correct Option: C

Explanation:

LCM of numeratorsHCF of denominators=LCM of 2,3,5HCF of 7,14,3=301=30${\displaystyle \frac{\text{LCM of numerators}}{\text{HCF of denominators}}=\frac{\text{LCM of 2,3,5}}{\text{HCF of 7,14,3}}=\frac{30}{1}=30}$

2. About the number of pairs which have 16 as their HCF and 136 as their LCM, the conclusion can be

a. only one such pair exists

b. only two such pairs exist

c. many such pairs exist

d. no such pair exists

Correct Option: D

Explanation:

HCF is always a factor of LCM. ie., HCF always divides LCM perfectly.

3. The HCF of two numbers is 12 and their difference is also 12. The numbers are

a. 66, 78

b. 94, 106

c. 70, 82

d. 84, 96

Correct Option: D

Explanation:

The difference of required numbers must be 12 and every number must be divisible by 12. Therefore, they are 84, 96.

4. The HCF of two numbers is 16 and their LCM is 160.  If one of the numbers is 32, then the other number is 

a. 48

b. 80

c. 96

d. 112

Correct Option:b

Explanation:

The number = HCF×LCMGiven number=16×16032=80

5. HCF of three numbers is 12. If they are in the ratio 1:2:3, then the numbers are

a. 12,24,36

b. 10,20,30

c. 5,10,15

d. 4,8,12

Correct Option: A

Explanation:

Let the numbers be a, 2a and 3a.

Then, their HCF = a  so a=12

The numbers are 12,24,36

LCM and HCF

1.About the number of pairs which have 16 as their HCF and 136 as their LCM, the conclusion can be
only one such pair exists only two such pair exists many such pairs exist no such pair exists

2. The HCF of two numbers is 12 and their difference is also 12. The numbers are
66,78 94,106 70,82 84,96

3.The HCF of two numbers is 16 and their LCM is 160. If one of the numbers is 32, then the other number is
48 80 96 112

4.HCF of three numbers is 12. If they are in the ratio 1:2:3, then the numbers are
12,24,36 10,20,30 5,10,15 4,8,12

5.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?
4 10 15 16

LCM or Least common factor and Highest common factor (HCF)or Greatest common divisor (GCD)

LCM is defined as the least number which is divisible by all the given divisors.  Take 4,6 as two divisors which divide 12, 24, 36... perfectly with no remainder.  So 12, 24, 36 are called common multiples of 4 and 6.  In other words, 4 and 6 are factors of all these number.  Of all these common multiples, 12 is the least number.  So we can say 12 is Least common multiple of all the given numbers or LCM of 4, 6.

Finding LCM: 

There are two ways to find LCM.  First one is division method, second one is Factorization method.  

1. Division Method: LCM of 15, 18, 27

In division method we have to continue the division until the numbers in the last row become co - primes with each other.  So LCM = 3 x 3 x 5 x 2 x 3 =270

2. Factorization Method: 

Here we can write all the given numbers in their prime factorization format.

15 = 3 x 5

18 = 2×32

27 = 33

Now take all primes number the given numbers and write their maximum powers. So LCM of 15, 18, 27 = 2×33×5= 270

Formula 1: If r is the remainder in each case when N is divided by x, y, z then the general format of the number is N= K x [LCM (x, y, z)] + r here K is a natural number

Example: A teacher when distributed certain number of chocolates to 4 children, 5 children, 7 children, left with 1 chocolate.  Find the least number of chocolates the teacher brought to the class

Ans:  N = K (LCM (4, 5, 7) + 1 = 140K + 1.  Where K = natural number.  When we substitute K = 1, we get the least number satisfies the condition. So minimum chocolates = 141

 

Formula 2: If x1,y1,z1 are the remainders when N is divided by x, y, z and x−x1=y−y1=z−z1=a then the general format of the number is given by N= K x [LCM (x, y, z)] - a

Example: When certain number of marbles are divided into groups of 4, one marble remained.  When the same number of marbles are divided into groups of 7 and 12 then 4, 9 marbles remained. If the total marbles are less than 10,000 then find the maximum possible number of marbles.

Ans:  In this case the difference between the remainders and divisors is constant.  i.e., 3. so  N = K (LCM (4, 7, 12) -  3 = 84K - 3.  Where K = natural number.  

But we know that 84K - 3 < 10,000 ⇒ 84 x 119  - 3 < 10,000 ⇒ 9996 - 3 = 9993