# What is the least common multiple (LCM) of 8, 12, and 18?

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#### Explanation

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Jul 6, 2016

Least Common Multiple is $72$

#### Explanation:

Multiples of $8$ are $\left\{8 , 16 , 24 , 32 , 40 , 48 , 56 , 64 , 72 , 80 , 88 , 96 , 104 , 112 , 120 , 128 , 136 , 144. \ldots . .\right\}$

Multiples of $12$ are $\left\{12 , 24 , 36 , 48 , 60 , 72 , 84 , 96 , 108 , 120 , 132 , 144. \ldots . .\right\}$

Multiples of $18$ are $\left\{18 , 36 , 54 , 72 , 90 , 108 , 126 , 144. \ldots . .\right\}$

Hence common multiples are $\left\{72 , 144 , \ldots \ldots .\right\}$

and Least Common Multiple is $72$

Another shorter way is to write numbers as multiplication of its prime factors

$8 = 2 \times 2 \times 2$ $\to$ $2$ comes three times

$12 = 2 \times 2 \times 3$ $\to$ $2$ comes twice and $3$ once.

$18 = 2 \times 3 \times 3$ $\to$ $2$ comes once and $3$ twice

So maximum times is three times for $2$ and two times for $3$.

Hence, Least Common Multiple is $2 \times 2 \times 2 \times 3 \times 3 = 72$.

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Tony B Share
Apr 9, 2018

Use Shwetyank's method in the class and an exam. The marking schema will be set up for that approach.

$\textcolor{b l u e}{\text{My alternative approach - helps with understanding.}}$

#### Explanation:

Target: LCM of 8,12 and 18.
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$\textcolor{m a \ge n t a}{\text{Once you are used to this it should look like}}$$\textcolor{m a \ge n t a}{\text{what follows (for these numbers).}}$

12=color(white)("d")8+4 color(white)("d") ->2xx4=color(white)(..)8" so "2xx12=24
24=18+6 color(white)("d")->3xx6=color(white)(.)18" so "color(white)(.)3xx24=72

$\text{LCM } = 72$

3 lines inclusive of the answer for 3 relationships

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$\textcolor{b l u e}{\text{Preamble about the method}}$

There are many ways of writing any particular value so we can make it look however we wish as long as final outcome is the correct value. Example: 6 can and may be written as $5 + 1$ if we so chose.

We can use this to our advantage.
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$\textcolor{b l u e}{\text{Answering the question}}$

$\textcolor{b r o w n}{\text{Dealing with the 8 and 12 part}}$

Write 12 as $8 + 4$

Now we start counting the 12's but make sure it encompasses the 8 + 4's. When we have accumulated enough 4's to make another eight we have found the least common factor of 8 and 12

$1 : \to \textcolor{g r e e n}{12} = \textcolor{red}{8} + 4$
$2 : \to \underline{\textcolor{g r e e n}{12}} = \textcolor{red}{8} + \underline{4 \leftarrow \text{ Add the 12's and the 4's}}$
color(white)("ddddd")color(green)(24) color(white)("ddd")ul(color(white)("ddd")color(red)(8))
color(white)("dddddddddddd")color(red)(24) larr" Adding all the 8's"

$2 \text{ of } 12 = 24$
3" of "color(white)("d")8=24

So LCF (for 8 and 12) is 24

So we have a fixed ratio of 8 and 12 at every sum of 24. Consequently any other product of factors will need to include some multiple of 24
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$\textcolor{b r o w n}{\text{Dealing with the 18 part}}$

As the value of 24 locks together the counts of 8 and 12 we carry that forward.

Write the 24 as $18 + 6$

We are now counting the 24's

$1 : \to \textcolor{g r e e n}{24} = \textcolor{red}{18} + 6$
$2 : \to \textcolor{g r e e n}{24} = \textcolor{red}{18} + 6$
$3 : \to \underline{\textcolor{g r e e n}{24}} = \textcolor{red}{18} + \underline{6 \leftarrow \text{ Add the 24's and 6's}}$
color(white)("ddddd") color(green)(72)color(white)("dddddd")ul(color(red)(18))larr" Add all the 18's"
$\textcolor{w h i t e}{\text{ddddddddddddd}} \textcolor{red}{72}$
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$\textcolor{m a \ge n t a}{\text{ LCM } = 72}$
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