- Books Name
- CBSE Class 6 Mathematics Book

- Publication
- Param Publication

- Course
- CBSE Class 6

- Subject
- Mathmatics

**Some problems on HCF and LCM**

**Ex. ** Find the greatest number which divides 43 and 91 leaving remainder 7 in each caes.

**Sol.** It is given that the required number when divides 43 and 91, the renainder is 7 in each case . This means that 43 – 7 = 36 and 91– 7 = 84 are completely divisible by required number .

Also, the required number is the greatest number satisfying the above property .

*∴* It is the HCF of 36 and 84.

36 = 2 × 2 × 3 × 3

84 = 2 × 2 × 3 × 7

*∴* Required HCF = 2 × 2 × 3 = 12

Hence, the required number = 12

**Ex. ** Find the largest number that will divide 20, 57 and 85 leaving remainders 2, 3 and 4 respecitively.

**Sol.** Clearly, the required number is the HCF of the number 20 – 2 = 18, 57 – 3 = 54 and 85 – 4 = 81.

18 = 2 × 3 × 3

54 = 2 × 3 × 3 × 3

81 = 3 × 3 × 3 × 3

*∴* Required HCF = 3 × 3 = 9

Hence, the required number = 9

**Relationship between HCF and LCM **

Let us take two numbers, say 16 and 24.

The HCFof 16 and 24 is 8.

The LCM of 16 and 24 is 48.

Since 8 is factor of 48, so we can say that HCF of the numbers is a factor of their LCM.

Product of HCF and LCM = 8 × 48 = 384

Product of Numbers = 16 x 24 = 384

So we can say that the product of two numbers is equal to the product of their HCF and LCM.

Let a and b are two numbers then

**a x b = HCF × LCM **

**Ex. ** The HCF of two number is 29 and their LCM is 1160. If one of the number is 145, find the other.

**Sol.** We know that

Product of the number = HCF × LCM

x × 145 = 29 × 1160

*∴* Required number = 232

**Ex. ** Find the least number which when divided by 6, 7, 8, 9 and 10 leaves reamainder 1.

**Sol.** As the remainder is same

Required number = LCM of divisors + Remainder

= LCM (6, 7, 8, 9, 10) +1

= 2520 + 1 = 2521

**Ex.** Find the least number which when decreased by 7 is exactly divisible by 12, 16, 18, 21 and 28.

LCM = 2 × 7 × 3 × 2 × 4 × 3 = 1008

Required number = 1008 + 7 = 1015.