Bea can run around the park in about 16minutes and her friend hillary can do it in about 20minutes if they both start the same time and continue running around the park, about how long will it take them to be next to each other again?

Interesting problem!

Bea runs around in 16 minutes, and Hillary in 20 minutes.They will meet again, AT the starting point, at times that are multiples of 16 or 20 minutes, which means we need to find the LCM of 16 and 20.We know that 16=2*2*2*2, and 20=(2*2)*5, so the LCM is 2*2*2*2*5=80 (I ignored the duplicated part, namely (2*2).
So in 80 minutes, they will meet again AT THE STARTING POINT.
Note:
this is a special case where the next time they meet will be at the starting point.  There are, however, cases where they may not meet at the starting point.  The general method is to consider how much faster Bea runs faster than Hillary PER MINUTE, and calculate accordingly.  Here's how.

Bea runs faster than Hillary by 1/16-1/20=1/80 laps per minute.  So it takes 1/(1/80)=80 minutes to catch up one whole lap.  Same answer as before, but in this case, they meet at the starting point.  That is not always the case.
Another example:The long hand of a clock runs one lap every 1 hour, and the short hand runs one lap every 12 hours.  Every hour, the long hand catches up by (1/1-1/12)=11/12 hours.
Time for long hand to catch up one whole lap with short hand is therefore 1/(11/12)=12/11 hours, or roughly one hour 5 minutes and 27 seconds.  We all know that they don't meet again at the starting point.  That will take 12 hours, which is the LCM of 1 and 12.

Ikr they be playing