## Least Common Multiple LCM

Least Common Multiple (LCM) of Numbers:

Multiples of 4 are: 4, 8, 12, 16, 20, 24, 28, 30, 32, 36 and so on endlessly.

Multiples of 6 are: 6, 12, 18, 24, 30, 36 and so on endlessly.

Now, multiples common to 4 and 6 are: 12, 24, 36 and so on endlessly.

But, among all these common multiples, 12 is the least one.

Therefore, 12 is called the Least Common Multiple i.e. L.C.M of 4 and 6.

How to Find LCM of Two or More Numbers:

Example 1:

Find the LCM of 4 and 6.

Solution:

4 = 22 and 6 = 2 × 3.

Now, the LCM of 4 and 6 is:

22 × 3 = 4 × 3 = 12.

Example 2:

Find the LCM of 8 and 9.

Solution:

8 = 23 and 9 = 32

Now, the LCM of 8 and 9 is:

23 × 32 = 8 × 9 = 72.

Since 8 and 9 are co-primes, i.e. numbers that do not have any common factor, except 1, their LCM is same as their product.

For example:

LCM of the two numbers 4 and 5 is 4 × 5 = 20 because the common factor of 4 and 5 is only 1.

Example 3:

Find the LCM of 30 and 36

Solution:

30 = 2 × 15 = 2 × 3 × 5, and

36 = 4 × 9 = 22 × 32.

Now, 22, 32 and 5 are the numbers that are highest powers of the factors: 2, 3 and 5 in 30 and 36.

Therefore, the LCM of 30 and 36 is product of the numbers that are highest powers of all the factors.

i.e. LCM of 30 and 36 = 22 × 32 × 5 = 4 × 9 × 5 = 180

2. How to Find HCF and LCM of Fractions:

Example1:

Find the HCF of the fractions: 2/3, 3/4, 4/5

Solution:

HCF of Numerators 2, 3 and 4 is 1.

LCM of denominators 3, 4 and 5 is 3 × 4 × 5 = 60

Therefore, HCF of 2/3, 3/4, 4/5 is 1/60

Example 2:

Find the HCF of the fractions: 8/9, 16/27, 64/81

Solution:

8 = 23 and 16 = 24 and 64 = 26

Therefore, HCF of 8, 16 and 64 is 23 i.e. 8, and

9 = 32, 27 = 33 and 81 = 34

Therefore, LCM of 9, 27 and 81 is 34 i.e. 81

Now, the HCF of the given fractions:

8/9, 16/27, 64/81 is:

(HCF of numerators)/ (LCM of denominators) = 8/81

Example 1:

Find the LCM of the fractions: 8/5, 7/4, 3/8

Solution:

Recall that LCM of numbers is their product when they do not have any common factor other 1.

Therefore, LCM of the numerators 8, 7 and 3 is:

8 × 7 × 3 = 168

HCF of the denominators 5, 4 and 8 is:

5 = 51, 4 = 22 and 8 = 23

Among 51, 22 and 23, there is no common factor.

Therefore, 1 is the common factor of the denominators 5, 4 and 8.

Therefore, LCM of the fractions: 8/5, 7/4, 3/8 is 168/1 = 168

Word Problems on LCM:

Example 1:

Find the least number which when divided by 3, 4 and 5 leaves a constant remainder 2 in each case.

Solution:

Formula:

First find the LCM of the three numbers 3, 4, and 5.

Since 3, 4, and 5 do not have any common factor, their LCM is their product.

Therefore, LCM of 3, 4, 5 = 3 × 4 × 5 = 60.

Now the required number is:

LCM of 3, 4, 5 + the constant remainder. i.e.

60 + 2 = 62

Verification:

Cross-check by dividing 62 by each of the given divisors: 3, 4 and 5.

You will note that 62 leaves a constant remainder 2 on being separately divided by each of the given divisors 3, 4 and 5.

Example 2:

Find a smallest number that is divisible by 20, 25 and 30. Also, the number is a perfect square.

Solution:

A number divisible by all the three numbers 20, 25 and 30 must their LCM.

So, the LCM of 20, 25 and 30:

20 = 4 × 5 = 22 × 5

25 = 52

30 = 2 × 15 = 2 × 3 × 5

Now, LCM of 20, 25 and 30 is product of numbers that are highest powers of the different factors in the three numbers, i.e.

22 × 3 × 52

But, the number must also be a perfect square.

To make the LCM 22 × 3 × 52 a number that is a perfect square, multiply it by 3.

So, the number is 22 × 32 × 52 = 4 × 9 × 25 = 900

Hence 900 is the perfect square that is exactly divisible by all the three numbers: 20, 25 and 30.

Example 3:

Find the minimum number of oranges in a bag that can be exactly divided among a group of 5 boys, a group of 7 girls and a group of 10 women?

Solution:

The required minimum number of oranges is the LCM of the three numbers:

5, 7 and 10.

Since 5 is a factor of 10, so find LCM of only 7 and 10.

Again, since 7 and 10 do not have any common factor, therefore, LCM of 7 and 10 is their product: 7 × 10 = 70

Example 4:

Two clocks ring an alarm at regular intervals of 5 mins and 8 mins. How many times do the two clocks ring alarms simultaneously between 5 am and 8 am, if

they both rung an alarm at 5 am?

Solution:

The two clocks will ring an alarm simultaneously at a time which is the LCM of their individual times of ringing alarms.

So, LCM of 5 and 8 is 5 × 8 = 40 mins.

Therefore, the two clocks will ring alarms simultaneously at intervals of 40 minutes.

Between 5 am and 8 am, there are 3 × 60 = 180 minutes.

So, 180mins/40mins = 180/40 = 4

How to find the number of numbers divisible by both 2 and 3