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
 A: I agree with previous answer that a=0 for any divosor. So only need to work with combinations of b,c and d. I will put now in this form for simplfy :
$$
F(b,c,d)=3^b 5^c 7^d
$$
What we need to get is a divisor which is $1\mod 4$. What happen if we multiply a ($1 \mod 4$) by 3. Then we will get ($3 \mod 4$). If we multiply a ($3 \mod 4$) we get ($1 \mod 4$). Same will happen if we multiply by 7.From 1 to 3, and from 3 to 1. If we multiply by 5 we get same $\mod 4$ : If we multiply ($1 \mod 4$) by 5 we get ($1 \mod 4$) and if we multiply ($3 \mod 4$) by 5 we get ($3 \mod 4$). 
What mean this ? We can use any c but b and d have a condition :
$$
b+d \equiv 0 \mod 2
$$
Maximal values of b and c are 3, but maximal value of d is 5. We also count 0, and in case b=0,c=0 and d=0 also would be count as a divisor (1).
For c, then we have 4 possibilities (0,1,2 and 3). For b and d, we put now (b,d) in order to count all possibilities : 
(0,0) (0,2) (0,4) 
(1,1) (1,3) (1,5) 
(2,0) (2,2) (2,4) 
(3,1) (3,3) (3,5) 
So in total we have 3*4=12 possibilities for b and d. So in total we have :
$$
4*12=48
$$
possibilities.(4 possibilities for c and 12 possibilities for b and d)
So we have 48 divisors. Daniel
A: 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.
A: 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.
