Use definition of divisibility to do this I know this is a very simple question that I think high school students would know. 
However, I need help in expressing it using the laws of divisibility
Two athletes run a circular track at a steady pace so that the first complete one round in 8 minutes and the second in 10 minutes. If they both start from the same spot at 4pm, when will be the first time they return to the start together? 
Its obvious its 40 minutes later (Simple LCM). But how to do the working involving the laws of divisibility? (i.e. a divides b. a divides bc, for a, b and c are any integers, etc)
 A: If you wish to use division, you might consider this :
Let $a$ denote the number of laps run by the first runner at 8 minutes per lap, and let $b$ denote the number of laps run by the second runner at 10 minutes per lap.
Then we require $8a = 10b$.  Divide both sides by $2$ to get $4a = 5b$, or $4a - 5b = 0$.  Since $4$ and $5$ have no common divisor, the least integer solution of this equation is $a=5$ and $b=4$, and we have $8 \times 5 = 10 \times 4 = 40$ minutes.
A: $\newcommand{\lcm}{\operatorname{lcm}}$
One commonplace way to do this involves prime factorizations:
$$
\lcm(10,8) = \lcm(2\cdot5,\quad 2\cdot2\cdot2) = 2\cdot2\cdot2\cdot5.
$$
The number you get has at least as many of each prime factor as the ones you started with, thus three $2$s because $8$ has three $2$s and the other one doesn't have more than three, and one $5$ becaue $10$ has one $5$ and the other one doesn't have more than that.
So for example
$$
\lcm(63,18) = \lcm(3\cdot3\cdot7,\quad 2\cdot3\cdot3) = 2\cdot3\cdot3\cdot7 = 126.
$$
There is also a way that requires no prime factorizations: First use Euclid's algorithm to find the $\gcd$, then multiply and divide, thus (but for efficiency, CANCEL BEFORE MULTIPLYING):
$$
\lcm(10,8) = \frac{10\cdot8}{\gcd(10,8)} = \frac{10\cdot8}{2} = 40.
$$
"Cancel before multiplying" means either cancel the $10$ with the $2$ to get $5\cdot8$ or cancel the $8$ with the $2$ to get $10\cdot4$.  This may seem unimportant when tiny numbers like this are involved, but it can make things much more efficient when big numbers are involved.
A: You're only using the definition of $\operatorname{lcm}(8,10)=40$. The Least Common Multiple of two numbers is a positive number such that it is a multiple of both numbers, and yet, is the smallest one. In other words, it satisfies the following three properties:
If $L=\operatorname{lcm}(a,b)$:
$1 - L > 0$
$2 - a \mid L$ and $b \mid L$
$3- a \mid L'$ and $b \mid L'$ $\implies$ $L \mid L'$.
Now, about your question, every time that the first or second runner completes a round it takes $8$ minutes and $10$ minutes respectively. So, to complete $k$ rounds, it takes them $8k$ minutes and $10k$ minutes respecitevely. Now, if we let $L$ represent the time that these two runners will be at the start point again then $8 \mid L$ and $10 \mid L$. But we're looking for the smallest such $L$, because we wish to know what is the first time that they return to the start together. That means that $L$ must be the $\operatorname{lcm}(8,10)$ which is $40$. So, the two runners will return to the start point together after $k=5$ rounds and $40$ minutes.
