Rick is thinking of a positive factor of $14$ and Steve is thinking of a positive factor of $42$.  If Rick and Steve are thinking of the same number, how many possible numbers could they be thinking of?
We will find the positive divisors of 14 by finding pairs that multiply to 14.   We begin our list as follows, $$1 \quad \underline{\hphantom{10}} \quad \dots \quad \underline{\hphantom{10}} \quad 14.$$  Checking $2$, we find that $2\cdot 7=14$, so our list becomes $$1 \quad 2 \quad \underline{\hphantom{10}} \quad \dots \quad \underline{\hphantom{10}} \quad 7 \quad 14.$$  Checking $3$, $4$, $5$, and $6$, we find that none of these are divisors of $14$, so our final list is $$1 \quad 2 \quad 7 \quad 14.$$  Next, we use the buddy method to determine the factors of $42$.  We begin our list as follows, $$1\quad \underline{\hphantom{10}} \quad \dots \quad \underline{\hphantom{10}} \quad 42.$$  Checking $2$, we find that $2\cdot 21=42$, so our list becomes $$1\quad 2 \quad \underline{\hphantom{10}} \quad \dots \quad \underline{\hphantom{10}} \quad 21 \quad 42.$$  Checking $3$, we find that $3\cdot 14=42$, so our list becomes $$1\quad 2 \quad 3 \quad \underline{\hphantom{10}} \quad \dots \quad \underline{\hphantom{10}} \quad 14 \quad 21 \quad 42.$$  Checking $4$ and $5$ we find that $4$ and $5$ are not divisors of $42$.  Checking $6$, we find that $6\cdot 7=42$, so our list becomes $$1\quad 2 \quad 3 \quad 6 \quad \underline{\hphantom{10}} \quad \dots \quad \underline{\hphantom{10}} \quad 7 \quad 14 \quad 21 \quad 42.$$  Since $7$ is already on our list, our final list is $$1\quad 2 \quad 3 \quad 6 \quad 7 \quad 14 \quad 21 \quad 42.$$  We compare our lists for the factors of $14$ and the factors of $42$ to see that the factors that $14$ and $42$ share are $1$, $2$, $7$, and $14$.  Therefore, Rick and Steve could be thinking of $\boxed{4}$ possible numbers.  Note that since $14$ is a factor of $42$, all of the factors of $14$ are also factors of $42$.