LCM Formula

Example 1:

Find the LCM of 12 and 15.

Solution:

Given, two numbers are a = 12 and b = 15.

The prime factorization of 12 and 15 is,

12 = 2 x 2 x 3

15 = 3 x 5

Since 3 is the only common factor in both the numbers, then

GCF(12, 15) = 3

Now, the LCM of 12 and 15 can be determined by using the formula,

\(LCM(a,~b)=\frac{\left | ~a.b~ \right |}{GCF(a,~b)}\)

\(LCM(12,~15)=\frac{\left | ~12\times 15~ \right |}{GCF(12,15)}\)

\(=\frac{\left | ~12\times 15~ \right |}{3}\)        [Put the values]

\(=\left | ~12\times 5~ \right |\)     [Dividing 15 by 3]

= 60

Therefore, the least common multiple of 12 and 15 is 60.

Example 2:

Determine the LCM of two numbers 150 and 30.

Solution:

Let a = 150 and b = 30

The prime factors for a and b are

150 = 2 x 3 x 5 x 5

30 = 2 x 3 x 5

As we know, the product of all the common factors gives the GCF of 30 and 150.

That is, 2 x 3 x 5 = 30

Now, the formula to find the lowest common multiple is,

\(LCM(a,~b)=\frac{\left | ~a.b~ \right |}{GCF(a,b)}\)

\(LCM(30,~150)=\frac{\left | ~30~\times~ 150~ \right |}{GCF(30,\ 150)}\)     [Put the values]

\(=\frac{\left | ~30~\times ~150 ~\right |}{30}\)                                  [Dividing 30 by 30]

= 150

Hence, the LCM of 150 and 30 is 150.

Example 3:

The bell at St. Albert’s School is programmed to ring every 50 minutes for students in grades 1 to 5. While another bell will ring once every hour for students in grades 6 to 10. How many times in the next 10 hours will both the bells ring simultaneously?

Solution:

According to the given information, let ‘a’ be the bell for the grades 1 to 5 that rings for every 50 mins.

And the bell ‘b’ is for grades 6 to 10 that rings for every 1 hr (60 mins).

In order to find how many times in the next 12 hours both the bells ring, we have to find the LCM of both the bells.

Then, let’s find the prime factors of 50 and 60.

50 = 2 x 5 x 5

60 = 2 x 2 x 3 x 5

The product of the common factors of 50 and 60 are 5 x 2 = 10.

Thus, 10 is the GCF of 50 and 60.

The LCM of 50 and 60 can be find with the formula,

\(LCM(a,~b)=\frac{\left | ~a.b~ \right |}{GCF(a,~b)}\)

\(LCM(50,~60)=\frac{\left | ~50\ \times\ 60~ \right |}{GCF(50,\ 60)}\)

\(=\frac{\left | ~50\ \times\ 60~ \right |}{10}\)     [Dividing 60 by 10]

\(=\left | ~50\ \times\ 6~ \right |\)

\(=300\)

Time taken for both the bells to ring together = 300 minutes

\(= \frac{300}{60}\)         [Convert minutes into hours]

\(= 5\) hours

So, the number of times both the bells ring together = \(\frac{10\ \text{hours}}{5\ \text{hours}}=2\)