Divisors and Decimals

Question

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.

Answer 1

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, $f(n)=2^n$.

Answer 2

Suppose $n$ is such that the decimal expansion of ${\large{\frac{1}{n}}}$ has exactly $2010$ digits after the decimal point.

Since the decimal expansion terminates, it follows that $n=2^a5^b$ where $a,b$ are nonnegative integers.

By hypothesis, $$\frac{1}{n}=\frac{x}{10^{2010}}$$ where $x$ is a positive integer.

Note that $x$ can't be a multiple of $10$, else by removing a common factor of $10$ from the numerator and denominator of $${\large{\frac{x}{10^{2010}}}}$$ it would follow that the decimal expansion of ${\large{\frac{1}{n}}}$ has less than $2010$ digits. \begin{align*} \text{Then}\;\;&\frac{1}{n}=\frac{x}{10^{2010}}\\[4pt] \iff\;&n=\frac{10^{2010}}{x}\\[4pt] \end{align*} hence $x$ divides $10^{2010}$.

Thus, the only possible prime factors of $x$ are $2$ and $5$.

Since $x$ is not a multiple of $10$, it follows that $x$ can't have both $2$ and $5$ as prime factors, hence $x$ must be a power of $2$ or a power of $5$.

To minimize $n$, we want to maximize $x$.

The largest power of $2$ which divides $10^{2010}$ is $2^{2010}$, and the largest power of $5$ which divides $10^{2010}$ is $5^{2010}$.

It follows that the largest possible value of $x$ is $5^{2010}$, hence the smallest possible value of $n$ is $2^{2010}$.

Finally, note that $2^{2010}$ has exactly $2011$ positive integer divisors, namely$2^0,2^1,2^2,...,2^{2010}$.