Number of divisors of the form $(4n+1)$ 
Find the number of divisors of $$2^2\cdot3^3\cdot5^3\cdot7^5$$ which are of the form $(4n+1)$  

I know how to find the total number of divisors. But, to find the number of divisors of the form $(4n+1)$, I'm thinking of listing down the divisors and then finding, but that'd be very tedious. Is there any elegant way to do this?  
Any help will be appreciated.
Thanks.
 A: Hint:
$(4n+3)(4k+3)=4(4nk+3(n+k)+2)+1$, applicable to $3, \; 7$; and
$(4n+1)^2=4(4n^2+2n)+1$, applicable to $5$.
A: Any positive divisor of $2^2\cdot 3^3\cdot 5^3\cdot 7^5$ of the form $4k+1$ is a number of the form:
$$ 3^a\cdot 5^b\cdot 7^c $$
with $0\leq a\leq 3,0\leq b\leq 3,0\leq c\leq 5$ and $a+c$ being even. There are:
$$ \frac{4\cdot 4\cdot 6}{2}=\color{red}{48} $$
ways to choose $a,b,c$ that way.
A: Number of divisors of $$N= 2^2\times3^3\times5^3\times7^5,$$which are of the form $4n+1$ exculuding 
$$\begin{align}1 &=\{ \text{number of terms in product}\}\\
&=(1+3^2)(1+5+5^2+5^3)(1+7^2+7^4)+\{\text{number of terms in product}\}\\
&= ( 3+3^3)(7+7^3+7^5)(1+5+5^2+5^3)-1\\
&= 2\times4\times3+2\times3\times4-1\\
&= 47
\end{align}$$
A: 
$$2^2\cdot3^3\cdot5^3\cdot7^5$$ , divisor of form $(4n+1)$
Main result used: [Modular product rule][1]

Note that $4n+1 \mod 4 = 1$, so we want the divisors to be  congurent to $1 \mod 4$.
For power of three, $ 3^2 \mod 4 =1$ and $3^3 \mod 4=3$
For power of five, $5 \mod 4 =1$ , $5^2 \mod 4 = 1$, $ 5^3 \mod 4 =1$
For power of seven, $7 \mod 4 =3$, $ 7^2 \mod 4= 1$...
For power of two,  $ 2 \mod 4 = 2$ and $2^2 \mod 4 = 0$
Let us see what powers don't effect the remainder.
Clearly any power of five is permissible as it doesn't effect the modular congruence,  even powers of seven and even powers of three are fine as they don't effect the remainder.
If we take odd power of seven and odd power of three, we get one $ 1 mod 4$
The above two conditions can be combined into the power of seven and power three always adding upto something even.
