jmc

We know that $\gcd (a,b)\cdot \mathrm{lcm}[a,b]=ab$ for all positive integers $a$ and $b$. Hence, in this case, $ab=200$. The prime factorization of 200 is $2^3\cdot 5^2$, so $a=2^p\cdot 5^q$ and $b=2^r\cdot 5^s$ for some nonnegative integers $p$, $q$, $r$, and $s$. Then $ab=2^{p+r}\cdot 5^{q+s}$. But $ab=200=2^3\cdot 5^2$, so $p+r=3$ and $q+s=2$.

We know that $\gcd (a,b)=2^{\min \{p,r\}}\cdot 5^{\min \{q,s\}}$. The possible pairs $(p,r)$ are $(0,3)$, $(1,2)$, $(2,1)$, and $(3,0)$, so the possible values of $\min \{p,r\}$ are 0 and 1. The possible pairs $(q,s)$ are $(0,2)$, $(1,1)$, and $(2,0)$, so the possible values of $\min \{q,s\}$ are 0 and 1.

Therefore, the possible values of $\gcd (a,b)$ are $2^0\cdot 5^0=1$, $2^1\cdot 5^0=2$, $2^0\cdot 5^1=5$, and $2^1\cdot 5^1=10$, for a total of $\boxed{4}$ possible values.