# How do you find the Least common multiple of 30ab^3, 20ab^3?

Dec 29, 2016

The LCM is $60 a {b}^{3}$.

#### Explanation:

The least common multiple (or LCM) of two numbers is the product of the largest amounts of each (prime) factor that appear in either number. In other words, it is the smallest value that we can guarantee will have both numbers as factors.

Step 1: Factor both numbers.

$30 a {b}^{3}$ has the factors $\left[\left(2 , 3 , 5 , a , {b}^{3}\right)\right]$.
$20 a {b}^{3}$ has the factors $\left[\left({2}^{2} , , 5 , a , {b}^{3}\right)\right]$.

Step 2: Compare the powers of each factor that appear in both numbers, and circle the one that's bigger.

The factor $2$ appears once in $30 a {b}^{3}$, and it appears twice in $20 a {b}^{3}$. Circle the ${2}^{2}$.

$\left[\begin{matrix}2 & 3 & 5 & a & {b}^{3} \\ \left({2}^{2}\right) & \null & 5 & a & {b}^{3}\end{matrix}\right]$

The factor $3$ appears once in $30 a {b}^{3}$, and not at all in $20 a {b}^{3}$. Circle the $3$.

$\left[\begin{matrix}2 & \left(3\right) & 5 & a & {b}^{3} \\ \left({2}^{2}\right) & \null & 5 & a & {b}^{3}\end{matrix}\right]$

The remaining three factors $\left(5 , a , {b}^{3}\right)$ appear the same number of times in both numbers. Circle either appearance of these factors.

$\left[\begin{matrix}2 & \left(3\right) & 5 & a & {b}^{3} \\ \left({2}^{2}\right) & \null & \left(5\right) & \left(a\right) & \left({b}^{3}\right)\end{matrix}\right]$

Step 3: Multiply these circled values together.

The circled values are $\left({2}^{2}\right) \left(3\right) \left(5\right) \left(a\right) \left({b}^{3}\right)$. The product of these values is

$60 a {b}^{3}$

This is our least common multiple.

Dec 30, 2016

The LCM is $60 a {b}^{3}$

#### Explanation:

$a {b}^{3}$ is in both of them so we only need to look at the numbers to determine LCM.

$\textcolor{b l u e}{\text{Condition 1}}$
The last digit in both 30 and 20 is 0. So the multiple must also end in 0.

$\textcolor{b l u e}{\text{Condition 2}}$
The first digits are 3 and 2. The 2 means that they both have to be factors of an even number. The closest even number they both divide exactly into is 6.

$\textcolor{b l u e}{\text{Combining condition 1 and 2}}$
6 put with 0 gives 60

Thus the LCM is $60 a {b}^{3}$