Number of Divisor of $2^{2}.3^{3}.5^{3}.7^{5}$ Number of Divisor of $2^{2}.3^{3}.5^{3}.7^{5}$ of the form $4n+1$ where n$\in \mathbb{N}$ is ........
My approach is to solve it using remainder theoran like putting $2^{a}\cdot 3^{b}\cdot 5^{c}\cdot 7^{d}$  put $a=0; b=0,2;c=0,1,2,3;d=0,2,4$ but not able to approach

Segregate the primes by class $\bmod 4$ since that is what we care about.  We can't have any factors of $2$ because then the number would either be $4n$ or $4n+2$.  Note that $3^2$ is of the form $4n+1$.  That is where your problem lies.
