Is there a positive integer with $2010$ distinct positive integer factors?

Question

Suppose that, $$f(n)=(\text{ the number of distinct positive integer factors of $n$ including $1$ and $n$ } )$$

Is there a positive integer $m$ such that $f(m)=2010$ ?

How can conclude about the existence of a positive integer $n$ such that $f(n)$ is equal to any desired positive integer?

Answer 1

Let $n = \prod p_i^{k_i}$. An integer factor of $n$ will be of the form $\prod p_i^{j_i}$ where $0 \le j_i \le k_i$. To "choose" a factor we must "choose" a power for each prime $p_i$ and for that prime we will have $k_i + 1$ choices.

So $f(n) = \prod(k_i + 1)$.

So the simplest answer would be $f(p^{2009}) = 2010$ for a prime $p$, but any $n = \prod p_i^{k_i}$ where $\prod(k_i + 1)= 2010$ will do. As $2010 = 2*3*5*67$, I think the smallest such number would be $n = 2^{66}*3^4*5^2*7$. At least I think that is the smallest such number.

Answer 2

Example $$m=2^{2009}$$

For all $n$, note that $2^{n-1}$ has $n$ distinct integer factors. Thus for all $n$, such $m$ exists that $f(m)=n$.