Factors of the form $4n+2$ $(n\geq 1)$ of a number N

Question

I need to find the total number of divisors of $N =2^53^45^{10}7^6$ that are of the form $4n+2$, $(n\geq1)$.

My attempt:-

$4n+2 = 2(2n+1)$

Clearly exponent of $2$ is only $1$, no other option is there, so $2^1$

$3,5,7$ are themselves of the form $(2n+1)$ so all of their exponents except 0 will satisfy the relation.

So I thought that the exponent of $3$ can be - $1,2,3,4$

Exponent of $5$ can be $1,2,3,4,5,6,7,8,9,10$

Exponent of $7$ can be $1,2,3,4,5,6$

So according to me the answer would be $4 \times 10 \times 6 = 240$

But I also noticed that $2 \times 7 \times 3^0$ would also satisfy this...so I have to take the 0th exponent of 3 afterall. So how would I do this?

Answer 1

Yes indeed. You have to take $0^{th}$ exponent in to account in all the cases. Or basically you have to find out the number of factors of $3^45^{10}7^6$ since $2$ is fixed anyways. That gives you $(4+1)(10+1)(6+1)=385$ [You know why there is $+1$?]

Subtract one to take off the case where you counted only $2\cdot3^05^07^0=2$