How many integers from 1 through 1000 are divisible by 3 and by at least one of 2,5,7, and 11? So I got 294 for this question, though I'm not a hundred percent sure but let me explain my process:
I did
to find all the multiples:
(floor of each)
1000/6 = 166 
1000/15 = 66
1000/21 = 47
1000/33 = 30
166+66+47+30 = 309
to eliminate the repeats:
(floor of each)
66/6 = 11 
47/6 = 7
30/6 = 5
47/15 = 3
30/15 = 2
30/21 = 1
11+7+5+3+2+1 = 29
to add in the repeats that were eliminated twice over:
(floor of each)
5/2 = 2
7/2 = 3 
11/2 = 5
7/5 = 1
11/5 = 2
11/7 = 1
2+3+5+1+2+1 = 14
309 - 29 + 14 = 294
 A: If you consider the range $3$ to $999$, divide by $3$, then we find that your question is equivalent to the following simpler question:

how many integers between $1$ and $333$ are divisible by at least one of $2$,  $5$, $7$, or $11$?

For this question we may use inclusion-exclusion. 

Multiples of $2,5,7,11$ with much overcounting:
$$\left\lfloor \frac{333}{2} \right\rfloor + \left\lfloor \frac{333}{5} \right\rfloor + \left\lfloor \frac{333}{7} \right\rfloor + \left\lfloor \frac{333}{11} \right\rfloor $$
Subtract these terms:
$$ \left\lfloor \frac{333}{10} \right\rfloor +  \left\lfloor \frac{333}{14} \right\rfloor + \left\lfloor \frac{333}{22} \right\rfloor + \left\lfloor \frac{333}{35} \right\rfloor + \left\lfloor \frac{333}{55} \right\rfloor + \left\lfloor \frac{333}{77} \right\rfloor $$
Add back these terms:
$$ \left\lfloor \frac{333}{70} \right\rfloor + \left\lfloor \frac{333}{110} \right\rfloor + \left\lfloor \frac{333}{154} \right\rfloor + \left\lfloor \frac{333}{385} \right\rfloor $$
Since $770>333$, there are no multiples of $770$ that we have to worry about having overcounted. 
