# What is the LCM of 36,8,72,12?

May 3, 2018

72

#### Explanation:

So $L C M$ means the lowest common multiple which in other words means what is the smallest number that is divisible by all the numbers you have present.

ie what is the smallest number which can be divided by $36 , 8 , 72 \mathmr{and} 12$ without any remainders?

So looking at the multiples of all the numbers you have

multiples of $36 : 36 , \textcolor{R e d}{72} , 108 , \ldots$
multiples of $8 : 8 , 16 , 24 , 32 , 40 , 48 , 56 , 64 , \textcolor{red}{72} , 80 , 88 , 96 , \ldots$
multiples of $72 : \textcolor{red}{72} , 144 , 216 , \ldots$
multiples of $12 : 12 , 24 , 36 , 48 , 60 , \textcolor{red}{72} , 84 , 96 , 108 , 120 , . .$

From above, you can clearly see that $\textcolor{red}{72}$ is the answer because it is the first common number that you can find in each of the multiples of $36 , 8 , 72 \mathmr{and} 12.$

May 5, 2018

$L C M = 2 \times 2 \times 2 \times 3 \times 3 = 72$

#### Explanation:

You can find the $L C M \mathmr{and} H C F$ of any set of numbers by breaking each one down into the product of its prime factors:

$\text{ "36 = 2xx2" } \times 3 \times 3$
$\text{ } 8 = 2 \times 2 \times 2$
$\text{ } 72 = 2 \times 2 \times 2 \times 3 \times 3$
" "12 = ul(2xx2" "xx3color(white)(xxxx))
$L C M = 2 \times 2 \times 2 \times 3 \times 3$

The $L C M$ must contain all the prime factors of all the numbers, but without unnecessary duplicates.

Both $8 \mathmr{and} 72$ have ${2}^{3}$
Both $36 \mathmr{and} 72$ have ${3}^{2}$

But we only need ${2}^{3} \mathmr{and} {3}^{2}$ to include all the numbers,