Here's the approach that I use when trying to find LCM.
$18 = 2\times3\times 3$
$12 = 2\times 2\times 3$
To be a multiple of 18 a number must have prime factors of at least one $2$ and two $3$.
To be a multiple of 12 a number must have prime factors of at least two $2$ and one $3$.
Hence the number with at least two $2$ and two $3$ prime factors will be the lowest number that is a common multiple of both $18$ and $12$.
Therefore, $2\times 2\times 3\times 3=36$ is the lowest common multiple of $12$ and $18$.
Would it be correct to 'expand' this a bit more like this in order to explain the above to someone:
To be a multiple of 18 a number must have prime factors of at least one $2$ and two $3$.
To be a multiple of 12 a number must have prime factors of at least two $2$ and one $3$.
Let's say the LCM of $12$ and $18$ is $x$
We know that a multiple of $12$ and $18$ is $12\times 18$
Which in prime factor form is: $2\times3\times 3 \times 2\times 2\times 3$
Let's say that this is the LCM we know of the two numbers so far, therefore:
$x =2\times3\times 3 \times 2\times 2\times 3$
We know that the minimum number of $2$'s required to be a multiple of both $12$ and $18$ is two. So let's cut off any extra $2$'s present in our LCM $x$:
$x =2\times3\times 3 \times 2\times$ 2 $\times 3$
We know that the minimum number of $3$'s required to be a multiple of both $12$ and $18$ is two. So let's cut off any extra $3$'s present in our LCM $x$:
$x =2\times3\times 3 \times 2\times$ 2 $\times$ 3
Hence our LCM $x$ has now been updated to:
$x =2\times3\times 3 \times 2$
Resulting in $x=2\times 2\times 3\times 3=36$