Can a multiple of $15$ and a multiple of $21$ differ by $1$? I know a solution to this question having to do with the fact that the $\gcd(15, 21) = 3$, so the answer is no.
But I can't figure out what is the reasoning behind this. Any help would be really appreciated! 

Suppose that $15m > 21n$. Then you are asking if:
$$15m - 21n= 1.$$
This corresponds to:
$$15m = 21n + 1.$$
The term on the left is divisible by $3$. What about the term on the right?
Well, $21n$ is divisible by $3$ too. But, if you add $1$ to $21n$, it can't be divisible by $3$ anymore. 
Actually, you have the left term which is divisible by $3$ and the right term which is not divisible by $3$.
For these reasons, you can't find those numbers. 
Similar arguments must be used to prove the same when $21n > 15m$. In this case, you should work on $21n = 15m + 1$.
