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

Guide:
Clearly $a=0$.
$$3 \equiv -1 \pmod 4$$
$$7 \equiv -1 \pmod 4$$
$$5 \equiv 1 \pmod 4$$
Hence we require $b+d$ to be an even number.
