Q:

Divisors and Decimals

For positive integers $n$, let $f(n)$ return the smallest positive integer $k$ such that $\frac1k$ has exactly $n$ digits after the decimal point. How many positive integer divisors does $f(2010)$ have?

I understand what the question is asking for, but I don't understand how to solve it. Here's what I have so far. The question is asking for a number $n$ that when taking the reciprocal of $n$ has 2010 digits after the decimal point. This means that $\frac1n$ is a terminating decimal. Given this piece of information, I know that $n$ can be expressed in the form $2^a\cdot5^b$. Help is greatly appreciated.

A:

Not only does $f(2010)=k=2^a5^b$, but each division by 10 adds one more digit (a zero) after the decimal point, and after all factors of 10 are used up, each further division by 2 or by 5 adds one more digit too. Therefore
$$2010=\max(a,b)$$
It then follows that the smallest such $k$ has $a=2010$ and $b=0$, i.e. $k=2^{2010}$. Its number of divisors is 2011 ($2^0,2^1,\dots,2^{2010}$), which is the final answer.
Indeed, $f(n)=2^n$.

