smc

Using the laws of logarithms, the given equation becomes $${\log }_{10}{2}^a+{\log }_{10}{3}^b+{\log }_{10}{5}^c+{\log }_{10}{7}^d=2005$$ $$\Rightarrow {\log }_{10}{2}^a\cdot 3^b\cdot 5^c\cdot 7^d=2005$$ $$\Rightarrow 2^a\cdot 3^b\cdot 5^c\cdot 7^d=10^{2005}$$ As $a,b,c,d$ must all be rational, and there are no powers of $3$ or $7$ in $10^{2005}$, $b=d=0$. Then $2^a\cdot 5^c=2^{2005}\cdot 5^{2005}\Rightarrow a=c=2005$. Only the four-tuple $(2005,0,2005,0)$ satisfies the equation, so the answer is $\boxed{1}\Rightarrow (\mathrm{B})$.