## Least Count Multiple (LCM) | How To Find LCM Of Two Numbers

The full form of LCM is the Least Common Multiple. LCM can be useful in solving many mathematical problems. In many competitive exams questions related to LCM are asked. It can be used before fractions can be added, subtracted or compared.

### Least Count Multiple

Lowest Common Multiple (LCM): The least or smallest common multiple of any two or more given natural numbers are termed as LCM.

Let’s find the LCM of 20 and 35 by the following steps

Step 1: First of all, in order to find the LCM of two numbers we will find the prime factors of each number.

20 = 2×2×5

35 = 5×7

Step 2: Now, we will multiply each factor with it’s greatest number of repetition in either number.

Step 3: If the same factor repeats more than once in both the numbers, then we’ll multiply the factor with it’s greatest number of repetition.

Here, 2- two-time occurred

5- one time occurred

7- one time occurred

2 × 2 × 5 × 7 = 140 = LCM

Problems

Example 1 7,9,49

Solutions:– According to above steps, list all prime factors of each number.

7:- 7

9:- 3 × 3

49:- 7 × 7

Now, you will have to multiply each factor with it’s greatest number of repetition in any of the numbers.

Here, 7 is two times occurred

3 is two times occurred

7 x 7 x 3 x 3= 441= LCM

9 has two 3s and 49 has two 7s, so we will multiply 3 and 7 both two times. That will give us 441, i.e, the smallest number that can be divided evenly by 3 and 7 both. Now check out your result by verifying that 441 can be divided evenly by 3 and 7 both.

Example 2: 14,40

Solutions :- According to above steps, list all the prime factors of each numbers.

14:- 2 × 7

40:- 2 × 2 × 2 × 5

Now you will have to multiply factor with it’s greatest number of repetition in any of the numbers.

Here 2 is three-time occurred

7 is one time occurred

5 is one time occurred

2 x 2 x 2 x 7 x 5 =280 = LCM

14 has one 2 and 40 has three 2s, so we’ll multiply three times 2 and now 14 has one 7 and 40 has one 5, so we’ll multiply 7 and 5 only one time. That will give us 280 i.e, the smallest number that can be divided by 2,7 and 5.

Now, check your result by verifying that 280 can be divided evenly by 2,7 and 5.