Integers divisible by 4 but not by 3 and 16 For $n \leq 1000$ I am interested in the integers who are divisible by 4 but not by 3 and 16. Say $a_i$ is the property that an integer is divisible by $i$. Inclusion-Exclusion gives us:
\begin{align}
N(a_{3}'a_{4}a_{16}') = N - N(a_3) - N(a_4') - N(a_{16}) + N(a_3a_4') + N(a_3a_{16}) + N(a_4'a_{16}) - N(a_{3}'a_{4}a_{16}') 
\end{align}
I can determine those first 4 terms correctly:
• N = 1000
• $N(a_3)= \lfloor \frac{1000}{3} \rfloor = 333$
• $N(a_4')= 1000 - \lfloor \frac{1000}{4} \rfloor = 750 $
• $N(a_{16})= \lfloor \frac{1000}{16} \rfloor = 62$ 
Now it is becoming harder to working things out. We can say for $N(a_3a_4')$ that those are all the numbers that are divisible by 3 but not by 4. We can again apply inclusion-exlusion on this term but it don't seem to help. How can I work this out?

$250$ numbers are divisible by $4$. Now use inclusion/exclusion principle:
$$250-\left\lfloor\frac{250}{3}\right\rfloor-\left\lfloor\frac{250}{4}\right\rfloor+\left\lfloor\frac{250}{3\times4}\right\rfloor=125$$
Please note that we use $4$ instead of $16$, because those $250$ numbers are not consecutive - they already contain only multiples of $4$, so we need to exclude only those that are multiples of yet "another $4$" (the credit for this fix goes to @AndréNicolas, as implied in the comment below).
