Note that the prime factorization of $210$ is $2\cdot 3\cdot 5\cdot 7$, and so the prime factorization of $210^3$ is $2^3\cdot 3^3\cdot 5^3\cdot 7^3$.

Given that $\gcd(a,b)=210$ and $\mathop{\text{lcm}}[a,b]=210^3$, we must have $a=2^k\cdot 3^\ell\cdot 5^m\cdot 7^n$ and $b=2^p\cdot 3^q\cdot 5^r\cdot 7^s$ where each of the ordered pairs $(k,p),(\ell,q),(m,r),(n,s)$ is either $(1,3)$ or $(3,1)$. Therefore, if we ignore the condition $a<b$, there are independently two choices for each of $k$, $\ell$, $m$, and $n$, and these choices determine both of the numbers $a$ and $b$. We have $2\cdot 2\cdot 2\cdot 2=16$ ways to make all four choices.

However, these $16$ sets of choices will generate each possible pair of values for $a$ and $b$ in both possible orders. Half of these choices will satisfy $a<b$ and half will satisfy $a>b$. So, imposing the condition $a<b$, we see that there are $\frac{16}{2}=\boxed{8}$ possible choices for $a$.
<|im_start|>system
You are a helpful assistant.<|im_end|>
<|im_start|>user
If $a$ and $b$ are positive integers such that $\gcd(a,b)=210$, $\mathop{\text{lcm}}[a,b]=210^3$, and $a<b$, how many possible values are there for $a$?
Please reason step by step, and put your final answer within \boxed{{}}.<|im_end|>
<|im_start|>assistant

user
8
If $a$ and $b$ are positive integers such that $\gcd(a,b)=210$, $\mathop{\text{lcm}}[a,b]=210^3$, and $a<b$, how many possible values are there for $a$?
test