jmc

The given equation is equivalent to $${\log }_{10}\left(2^a\cdot 3^b\cdot 5^c\cdot 7^d\right)=2005,\text{so}{2}^a\cdot 3^b\cdot 5^c\cdot 7^d=10^{2005}=2^{2005}\cdot 5^{2005}.$$Express $a$, $b$, $c$ and $d$ as simplified fractions and let $M$ be the least common multiple of their denominators. It follows that $$2^{Ma}\cdot 3^{Mb}\cdot 5^{Mc}\cdot 7^{Md}=2^{2005M}\cdot 5^{2005M}.$$Since the exponents are all integers, the Fundamental Theorem of Arithmetic implies that $$Ma=2005M,Mb=0,Mc=2005M,\text{and}Md=0.$$Hence the only solution is $(a,b,c,d)=(2005,0,2005,0)$, and the answer is $\boxed{1}$.