LCM and HCF - Solved Examples

11.The total number of prime factors of the product (8)20×(15)24×(7)15$(8{)}^{20}\times (15{)}^{24}\times (7{)}^{15}$ is

a. 59

b. 98

c. 123

d. 4

Correct Option: D

Explanation:

The prime numbers are 2,3,5,17 in the expression.  The expression can be written as (23)20×(3×5)24×(17)15⇒260×324×524×1715$(2^3{)}^{{}^{20}}\times (3\times 5{)}^{24}\times (17{)}^{15}\Rightarrow 2^{60}\times 3^{24}\times 5^{24}\times {17}^{15}$

So number of prime factors are 4. i.e., 2, 3, 5, 17

12. The HCF and LCM of two numbers are 44 and 264 respectively. If the first number is divisible by 3, then the first number is

a. 264

b. 132

c. Both a and b

d. 33

Correct Option: C

Explanation:

Let the numbers are ah, bh respectively.  Here h is HCF of two numbers.  (obviously a, b are coprimes i.e., HCF (a, b) = 1)

Given that HCF = h = 44 and LCM = abh = 264

Dividing LCM by HCF we get ab = 6.

ab can be written as 1 x 6, 2 x 3, 3 x 2, 6 x 1.

But given that the first number is divisible by 3. So only two options possible for A. 3 x 44, 6 x 44. So option C is correct

13. What least number must be subtracted from 1294 so that the remainder when divided 9, 11, 13 will leave in each case the same remainder 6 ?

a. 0

b. 1
c. 2
d. 3

Correct Option: B

Explanation:

LCM of 9,11,13 is 1287. Dividing 1294 with 1287, the remainder will be 7, to get remainder 6, 1 is to be deducted from 1294 so that 1293 when divided by 9,11,13 leaves 6 as remainder.

14. The least number which is divisible by 12, 15, 20 and is a perfect square, is

a. 400

b. 900
c. 1600

d. 3600

Correct Option: b

Explanation: 

LCM = 5 × 3 × 2² = 60
To make this number as a perfect square, we have to multiply this number by 5 × 3

The number is 60 × 15= 900

15. The least perfect square number which is divisible by 3,4,5,6 and 8 is

a. 900

b. 1200

c. 2500

d. 3600

Correct Option: D

Explanation: 

LCM = 2×2×2×3×5=23×3×5$2\times 2\times 2\times 3\times 5=2^3\times 3\times 5$

But the least perfect square is = 23×3×5×(2×3×5)=24×32×52=3600$2^3\times 3\times 5\times (2\times 3\times 5)=2^4\times 3^2\times 5^2=3600$  as the perfect squares have their powers even.

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