## Find The HCF And LCM using Prime Factorisation Method

**Relation between two numbers and their HCF and LCM **

Consider two numbers 18 and 24.

Prime factorisation of 18 = 2 × 3 × 3

Prime factorisation of 24 = 2 × 2 × 2 × 3

So, HCF = 2 × 3 = 6

LCM = 2 × 2 × 2 × 3 × 3 = 72

Product of HCF and LCM = 6 × 72 = 432

Product of given numbers = 18 × 24 = 432

The product of LCM and HCF of two natural numbers is equal to the product of the given natural numbers.

∴ Product of given numbers = HCF × LCM of given numbers

For any two positive integers:

Their LCM. × their HCF. = Product of the number

(i) LCM = \(\frac{\text{Product of the numbers}}{\text{HCF}}\)

(ii) HCF = \(\frac{\text{Product of the numbers}}{\text{LCM}}\)

(iii) One number = \(\frac{\text{H}\text{.C}\text{.F}\text{. }\!\!\times\!\!\text{ L}\text{.C}\text{.M}\text{.}}{\text{Other number}}\)

## Finding HCF And LCM using Prime Factorisation Method **Example Problems With Solutions**

Find the L.C.M. and H.C.F. of the following pairs of integers by applying the Fundamental theorem of Arithmetic method i.e., using the prime factorisation method.

**Example 1: **26 and 91

**Sol.** Since, 26 = 2 × 13 and, 91 = 7 × 13

**L.C.M.** = Product of each prime factor with highest powers. = 2 × 13 × 7 = 182.

i.e., **L.C.M.** (26, 91) = 182.

**H.C.F. **= Product of common prime factors with lowest powers. = 13.

i.e., **H.C.F** (26, 91) = 13.

Product of given two numbers = 26 × 91 = 2366

and, product of their **L.C.M.** and **H.C.F.** = 182 × 13 = 2366

Product of L.C.M and H.C.F of two given numbers = Product of the given numbers

**Example 2: **1296 and 2520

**Sol.** Since, 1296 = 2 × 2 × 2 × 2 × 3 × 3 × 3 × 3 = 2^{4} × 3^{4}^{ }

and, 2520 = 2 × 2 × 2 × 3 × 3 × 5 × 7 = 2^{3} × 3^{2} × 5 × 7

**L.C.M.** = Product of each prime factor with highest powers

= 2^{4} × 3^{4} × 5 × 7 = **45,360 **

i.e., **L.C.M.** (1296, 2520) = 45,360

**H.C.F. **= Product of common prime factors with lowest powers = 2^{3} × 3^{2} = 8 × 9 = 72

i.e., H.C.F. (1296, 2520) = **72.**

Product of given two numbers = 1296 × 2520 = 3265920

and, product of their **L.C.M.** and **H.C.F.** = 45360 × 72 = 3265920

L.C.M. (1296, 2520) × H.C.F. (1296, 2520)

= 1296 × 2520 = 3265920

**Example 3: **17 and 25

**Sol.** Since, 17 = 17

and, 25 = 5 × 5 = 5^{2}

**L.C.M.** = 17 × 5^{2} = 17 × 25 = 425

and, **H.C.F.** = Product of common prime factors with lowest powers = 1, as given numbers do not have any common prime factor.

The given numbers 17 and 25 do not have any common prime factor. Such numbers are called co-prime numbers and their H.C.F. is always equal to 1 (one), whereas their L.C.M. is equal to the product of the numbers.

But in case of two co-prime numbers also, the product of the numbers is always equal to the product of their L.C.M. and their H.C.F.

As, in case of co-prime numbers 17 and 25;

H.C.F. = 1; L.C.M. = 17 × 25 = 425;

product of numbers = 17 × 25 = 425

and product of their H.C.F. and L.C.M. = 1 × 425 = 425.

**Example 4: **Given that H.C.F. (306, 657) = 9, find L.C.M. (306, 657)

**Sol.** H.C.F. (306, 657) = 9 means H.C.F. of

306 and 657 = 9

Required L.C.M. (306, 657) means required L.C.M. of 306 and 657.

For any two positive integers;

their L.C.M. = \(\frac{\text{Product of the numbers}}{\text{H}\text{.C}\text{.F}\text{.}}\)

i.e., L.C.M. (306, 657) = \(\frac{306\times 657}{9}\) = 22,338.

**Example 5:** Given that L.C.M. (150, 100) = 300, find H.C.F. (150, 100)

**Sol.** L.C.M. (150, 100) = 300

⇒ L.C.M. of 150 and 100 = 300

Since, the product of number 150 and 100

= 150 × 100

And, we know :

H.C.F. (150, 100) = \(\frac{\text{Product of 150 and 100}}{L.C.M\text{.(150,100)}}\)

= \(\frac{150\times 100}{300}\) = 50.

**Example 6: **The H.C.F. and L.C.M. of two numbers are 12 and 240 respectively. If one of these numbers is 48; find the other numbers.

**Sol.** Since, the product of two numbers

= Their H.C.F. × Their L.C.M.

⇒ One no. × other no. = H.C.F. × L.C.M.

⇒ Other no. = \(\frac{12\times 240}{48}\) = 60.

**Example 7:** Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 + 5 are composite numbers.

**Sol.** Since,

7 × 11 × 13 + 13 = 13 × (7 × 11 + 1)

= 13 × 78 = 13 × 13 × 3 × 2;

that is, the given number has more than two factors and it is a composite number.

Similarly, 7 × 6 × 5 × 4 × 3 + 5

= 5 × (7 × 6 × 4 × 3 + 1)

= 5 × 505 = 5 × 5 × 101

∴ The given no. is a composite number.