<s>[INST] Number of Divisors of $N = 3^55^77^9$ of the form $4n+1$ I need to find the number of divisors of $N = 3^55^77^9$ that are of the form $4n+1$, $n\geq 0$.
My try:
I noticed that $5$ itself is a number of the form $4n+1$ so all of its power satisfy the required condition, so number for the exponent of $5$ will be $7+1 = 8$.
Also, even powers of $3$ and $7$ satisfy the required condition. 
But, I also noticed that odd powers of $3$ and $7$ multiplied together also satisfy the given condition as $(4m+3)(4k+3) = 4q+1$. How would I take that into account?
 [/INST] $N = 3^55^77^9$  gives $(7+1)((3)(5)+(3)(5)) =240$ the first $(3)(5)$ deals with $\{0,2,4\}$ and $\{0,2,4,6,8\}$ the second with $\{1,3,5\}$and $\{1,3,5,7,9\}$
</s>