A simple problem on Finding factors

Question

There's a question wherein we have to find the number of factors of $480$ of the form $8n+4$ where $n \geq 0$. Now, the prime factorisation of $480$ gives $2^5 \times 3 \times 5$.
This is to be of the form $8n+4$, i.e. $4(2n+1)= 4x$ (where $x$ is some odd number). If we equate $4kx$ with $32 \times 3 \times 5$ (k is $\frac{480}{x}$) we obtain the expression $kx = 8 \times 3 \times 5$.
And as $x$ is an odd number, $8$ can't be one of the prime factors of $x$. So $x$ can either be $3, 5$ or $15$. The corresponding values of $k$ would be $160, 96$ and $32$. Therefore, there can be three such values.
But trial and error gave me four values- $4, 12, 20$ and $60$. Where am I going wrong?

Answer 1

A full answer would be as follows. We require:- $$4(2n+1)=2^a3^b5^c.$$ Therefore $a$ is precisely $2$ and we have four possibilities given by $$(b,c)\in (0,0),(0,1),(1,0),(1,1).$$ That gives you the four numbers obtained by trial and error.