jmc

We will try to prove that $f(n)=2^n$. Given that $f(n)=k$, we know that $\frac{1}{k}$ has exactly $n$ digits after the decimal point. If we multiply $\frac{1}{k}$ by $10^n$, then all the digits are shifted $n$ places to the left, so we should end up with an integer that is not divisible by 10. Therefore, we want to find the smallest integer $k$ that divides $10^n$ and leaves a quotient that is not divisible by 10. If we let $k=2^n$, then the quotient is $5^n$, which is odd and therefore not divisible by 10. For any integer smaller than $2^n$, the maximum power of 2 that can divide such an integer is $2^{n-1}$, so there remains at least one power of two that combines with a power of five to form an integer divisible by 10. Therefore, we have proved that $f(n)=2^n$.

As a result, we can now conclude that $f(2010)=2^{2010}$. The only integers that can divide $2^{2010}$ are $2^x$, for $0\le x\le 2010$. There are $\boxed{2011}$ such integers.