I need to apply the Principle of Inclusion-Exclusion here:
Hence I get $$|A_8| + |A_5| + |\neg A_6| - |A_8 \cap A_5| - |A_8 \cap \neg A_6| - |\neg A_6 \cap A_5| + |A_8 \cap A_5 \cap \neg A_6|$$
I struggle to find the $\cap$ of $A_8$ and $A_5$ with $\neg A_6$
$A_8$ : numbers divisible by 8, $A_5$ : numbers divisible by 5 so $A_8 \cap A_5$ = $A_40$.
$A_{40} \cap \not A_6 = $ numbers divisible by 40 and not by 6. That's 2 out of 3 multiples of 40. As you are counting from 0 to 2400 you do the maths ;)