How many positive divisors of $2004^{2004}$ are divisible by exactly 2004 positive integers?

Question

This would mean that the the divisor would have exponents that when one is added to them, they would multiply to 8016. I have trouble finding those numbers.

Answer 1

As $2004=2^2\cdot 3\cdot 167$, the candidate divisors are of the form $n=2^a3^b167^c$. Such $n$ has $(a+1)(b+1)(c+1)$ positive divisors. Any factorization of $2004=uvw$ into three positive divisors gives us a solution $2^{u-1}3^{v-1}167^{w-1}$ (because each of $u,v,w$ will be $\le 2004$ and all prime exponents in $2004^{2004}$ are large enough). The prime $167$ will occur in exactly one of the factor, same for $3$. For the $2$'s we either have exactly one of $u,v,w$ is odd, or exactly one of $u,v,w$ is even. In total, we find $3\cdot 3\cdot (3+3)=54$ divisors.

Answer 2

$2004 = 2^2*3*167$

So $2004^{2004}= 2^{4008}3^{2004}167^{2004}$

So if $k|2004^{2004}$ then $k = 2^a3^b167^c$ where $0\le a \le 6012;0\le b\le 2004; 0\le c \le 2004$.

There are $4009$ choices for $a$ and $2005$ choices for $b$ and $c$. SO there are $4009*2005*2005$ divisors of $2004^{2004}$.

How many divisors does $k$ have?

The same argument:

If $m|k$ then $m = 2^x3^y167^x$ where $0\le x \le a$ and $0\le y\le b$ and $0\le y\le c$.

There are $(a+1)(b+1)(c+1)$ ways to choose these $x,y,z$ so $k$ has $(a+1)(b+1)(c+1)$ divisors.

So if $k$ has exactly $2004$ divisors then:

i) $(a+1)(b+1)(c+1) = 2004$

ii) $a \le 4008; b,c\le 2004$

How many ways are there to do this?

$(a+1)(b+1)(c+1) = 2^2*3*167$.

Basically $a+1 = 2^{m_a}3^{m_b}167^{m_c}; m_a\le 2;m_b \le 1; m_c \le 1$

while $b+1 = 2^{n_a}3^{n_b}167^{n_c}; n_a \le 2-m_a; n_b\le 1-m_b; n_c \le 1-m_b$.

And $c+1$ is $2^{2-m_a - n_a}3^{1 - m_b - n_b}167^{1-m_c-n_c}$.

How many ways are there to choose them $m_a, n_a,m_b, n_b,m_c,n_c$?

$(m_a,n_a)$ may be $(2,0), (1,0),(1,1),(0,2),(0,1),(0,0)$. $6$ choices.

$(m_b,n_b)$ may be $(1,0), (0,1), (0,0)$. $3$ choices.

Same for choicing $(m_c, n_c)$.

So there are $6*3*3 = 54$ such factors.

They are $2^a3^b167^c$ where $(a+1)(b+1)(c+1) = 2004$