Considering the set of integers between 77-999, inclusive, how many are divisible by no less than three of the numbers 2, 3, 6 and 8?

Following the ideas of Robert, I'll provide a solution. Consider the list $\lbrace 77,78,\dots,1000\rbrace$. Say a number is "good" if it is divisible by at least three of the numbers $2,3,6,8$. If $2$, or $3$, does not divide a number, neither does $6$, so such a number cannot be good, as there are at least two numbers from our list not dividing. Thus, we are only interested in numbers divisible by $2$, and by $3$ - but such numbers are also divisible by $6$, so three of our numbers do divide. We have thus characterized the good numbers completely as the numbers divisible by $6$, i.e. set $\lbrace 78,84,\dots,996\rbrace$. This is every sixth number in our set, i.e. we get $\lfloor \frac{1000-76}{6}\rfloor = 154$ as our answer.

“No less than three” means the same as “at least three”. So a number that meets your inclusion criterion must

• be divisible by three of the numbers, or
• be divisible by all four of the numbers (or equivalently, be divisible by their lowest common multiple, which is 24).