Here, we want to get the number which divides all numbers that are evenly divisible by 6 and 14
Looking at both numbers, we can see that at any point in time we can have;
[tex]\begin{gathered} 6\text{ = 2}\times3 \\ 14=2\times7 \end{gathered}[/tex]we can have 2 in two times and 3 in a single time and 7 in a single time. All we have to do is to multiply 3 and 7 which gives 21 (we simply cancel out the two on both ends)
This mean at any point in time, every number evenly divisible by 6 and 14 is also evenly divisble by 21
