number of divisors of $2^23^35^57^411^3$ which are is in the form of $6k+1, k\geq 0$ and $k\in \mathbb{Z}$ 
Total number of divisors of $2^23^35^57^411^3$ which are is in the form of $6k+1, k\geq 0$ and $k\in \mathbb{Z}$ 

$\bf{Attempt}$ writting $1,3,3^2,3^3$ as $6k+1$ or $6k+3$
same way $1,5,5^2,5^3,5^4,5^5$  as $6k+1$ or $6k+4$ 
same way $1,7,7^2,7^3,7^4$  as $6k+1$ 
same way $1,11,11^2,11^3$ as $6k+1$ or $6k+5$
could some help me how can i solve my question, thanks 
 A: Well, we can get $1 = 6k+1$ for $k=0$ off the bat, so that's nice. We can also ignore the $2$s and $3$s (we may do so by arguing that no divisor of $2^{2}3^{3}5^{5}7^{4}11^{3}$ with $2$ or $3$ as a factor will be congruent to $1$ mod $6$). Fix $a \in \mathbb{N}$. Since $7 = 6k+1$ for $k=1$, we clearly have $7^{a} \equiv 1 \pmod{6}$ for any $a$. For odd $a$, we can show that $$5^{a} \equiv 5 \pmod{6} \hspace{0.4cm} \text{and} \hspace{0.4cm} 11^{a} \equiv 5 \pmod{6}.$$
For even $a$, we may also show that
$$5^{a} \equiv 1 \pmod{6} \hspace{0.4cm} \text{and} \hspace{0.4cm} 11^{a} \equiv 1 \pmod{6}.$$
We should now make use of what we know about modular multiplication and the fact that $5^{2} \equiv 1 \pmod{6}$. This boils down to the question: how many combinations of $5^{i}7^{j}11^{k}$ (for $i,j,k \in \mathbb{N} \cup \lbrace 0 \rbrace$ and $i \leq 5$, $j \leq 4$, $k \leq 3$) can you make such that both $i$ and $k$ are even or odd?
A: Forget about the 2s and 3s and just consider factors of
$5^57^411^3$. How many of them are there? They are all of the form
$6k\pm 1$. One would guess about half of them are $6k+1$.
Each is $a11^j$ where $a\mid 5^57^4$ and $j\in\{0,1,2,3\}$. If $a\equiv1\pmod 6$ then $a11^j\equiv1\pmod 6$ for $j\in\{0,2\}$ and
$a11^j\equiv5\pmod 6$ for $j\in\{1,3\}$. For each of these $a$,
half the admissible $a11^j$ are $6k+1$. What about if $a\equiv5\pmod 6$?
