Find the number of those $[a]\in \langle[5]\rangle$ such that $\gcd(a,6)=1$

Question

Consider the finite cyclic group $\mathbb Z_{150}$. Consider the cyclic subgroup generated by $[5]$ which is $\langle[5]\rangle=\{[5],[10],[15],\ldots ,[145],[0]\}$ . Find the number of those $[a]\in \langle[5]\rangle$ such that $\gcd(a,6)=1$

My try

I know $150=2.3.5^2$. Now if I write in explicitly in hand, I found set of all $a$ such that $\gcd(a,6)=1$ to be the following set:

$\{5,25,35,55,65,85,95,115,125,145\}$.

Is there any explicit formula to calculate the above?

I will be grateful for some help.

In general my question is :

If $n=\prod_{i=1}^k p_i^{\alpha_i}$ what will be the number of those $[a]\in \langle p_k\rangle $ such that $\gcd(a,p_1p_2\cdots p_{k-1})=1$?

Answer 1

$[a]\in \langle[5]\rangle$ if and only if $a=5n$ for some integer, $1 \le n \le 29$.

$(a,6)=1 \iff (5n, 6)=1 \iff 2 \not \mid n \vee 3 \not \mid n$

There are $14$ multiples of $2$ between $1$ and $29$.

There are $9$ multiples of $3$ between $1$ and $29$.

There are $4$ multiples of $6$ (both $2$ and $3$) between $1$ and $29$.

Hence there are $14 + 9 - 4 = 19$ multiples of $2$ or $3$ between $1$ and $29$.

Hence there are $10 = 29-19$ elements $[a]\in \langle[5]\rangle$ such that $(a,6) = 1$.

Answer 2

The congruence system: $\,a\equiv 0\pmod{\!5},\,\ \overbrace{a\equiv \color{#0a0}{-1},\,\color{#90f}{1}\pmod{\!6}}^{\textstyle \gcd(a,6)=1}\ $ has $\,\color{#c00}2\,$ solutions $\,\color{#0a0}5,\color{#90f}{25}\pmod{\! 30}\,$ by CRT, which lift to $\,\color{#c00}2(150/30) = 10\,$ solutions $\!\bmod 150,\,$ viz $\,\{\color{#0a0}5,\color{#90f}{25}\}+30k,\ k = 0,\ldots, 4,\,$ since $\,5n\bmod 150 = 5(n\bmod 30)\,$ by the $\!\bmod\!$ Distributive Law.