Number of integers $x$ between $1$ and $95$ such that $96$ divides $60x$

Question

Question

Number of integers $x$ between $1$ and $95$ such that $96$ divides $60x$ is -:

$A$. $0$

$B$. $7$

$C$. $8$

$D$. $11$

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My Approach

For $a$ divides $b$, we can write as-:

$a \,\, |\,\,b \Rightarrow b=a \times c $,for some integer $c$

so

$96 \,\, |\,\,60x \Rightarrow 60x=96 \times c $

now putting option one by one

$x=7,$

$60 \times 7=96 \times c $ $c=\frac{420}{96}=4.375$ hence not possible as $c$ is an integer.

$x=8,$

$60 \times 8=96 \times c $ $c=\frac{480}{96}=5$ possible as $c$ is an integer.

$x=11,$

$60 \times 11=96 \times c $ $c=\frac{660}{96}=6.875$

not possible as $c$ is an integer.

hence $8$ should be the answer .is my reasoning correct? Please help

Answer 1

$60x$ is divisible by $96$ $\Rightarrow \frac{60x}{96}$ is a natural number

$\Rightarrow \frac{12\times 5x}{12 \times 8}$ is a natural number

$\Rightarrow \frac{5x}{8}$ is a natural number

$\Rightarrow x$ is divisible by $8$ because $GCD(5;8)=1$.

From $1$ to $95$, there are $11$ numbers satisfy this: $8;16;24;32;40;48;56;64;72;80;88$.