2-1-0......boom

$210=2*3*5*7$

$$

The number of divisors $d$ of a number $N=p^{\alpha_1}_1*p^{\alpha_2}_2*........*p^{\alpha_{n-1}}_{n-1}*p^{\alpha_n}_n$ is $d=(\alpha_1+1)*(\alpha_2+1)*........*(\alpha_{n-1}+1)*(\alpha_n+1)$ with $p_i$ prime factors

$$

A number multiple of $210$ is in the form $2^\alpha*3^\beta*5^\gamma*7^\delta*k$ . In this case $k=1$ because of the exponent 210 which has 4 divisors.

$$

Now it's necessary that $\alpha=(2-1)$ or $(3-1)$ or $(5-1)$ or $(7-1)$ $\Rightarrow$ $\alpha=1$ or $2$ or $4$ or $6$ . The same is for $\beta$ , $\gamma$ and $\delta$ .

$$

In conclusion $\alpha$ has 4 possibilities, $\beta$ has 3 possibilities, $\gamma$ has 2 possibilities and $\delta$ has 1 possibility. Total numbers are $1*2*3*4=\boxed{24}$