How many positive integers not exceeding $2001$ are multiples of $3$ or $4$ but not $5$?
$\textbf{(A) }768 \qquad \textbf{(B) }801 \qquad \textbf{(C) }934 \qquad \textbf{(D) }1067 \qquad \textbf{(E) }1167$
Solution 1
Out of the numbers $1$ to $12$ four are divisible by $3$ and three by $4$, counting $12$ twice.
Hence $6$ out of these $12$ numbers are multiples of $3$ or $4$. 
The same is obviously true for the numbers $12k+1$ to $12k+12$ for any positive integer $k$.
Hence out of the numbers $1$ to $60=5\cdot 12$ there are $5\cdot 6=30$ numbers that are divisible by $3$ or $4$.
Out of these $30$, the numbers $15$, $20$, $30$, $40$, $45$ and $60$ are divisible by $5$.
Therefore in the set $\{1,\dots,60\}$ there are precisely $30-6=24$ numbers that satisfy all criteria from the problem statement.
Again, the same is obviously true for the set $\{60k+1,\dots,60k+60\}$ for any positive integer $k$.
We have $1980/60 = 33$, hence there are $24\cdot 33 = 792$ good numbers among the numbers $1$ to $1980$. At this point we already know that the only answer that is still possible is $\boxed{\textbf{(B)}}$, as we only have $20$ numbers left.
By examining the remaining $20$ by hand we can easily find out that exactly $9$ of them match all the criteria, giving us $792+9=\boxed{\textbf{(B) }801}$ good numbers.
This is correct.

Solution 2
We can solve this problem by finding the cases where the number is divisible by $3$ or $4$, then subtract from the cases where none of those cases divide $5$. To solve the ways the numbers divide $3$ or $4$ we find the cases where a number is divisible by $3$ and $4$ as separate cases. We apply the floor function to every case to get $\left\lfloor \frac{2001}{3} \right\rfloor$, $\left\lfloor \frac{2001}{4} \right\rfloor$, and $\left\lfloor \frac{2001}{12} \right\rfloor$. The first two floor functions were for calculating the number of individual cases for $3$ and $4$. The third case was to find any overlapping numbers. The numbers were $667$, $500$, and $166$, respectively. We add the first two terms and subtract the third to get $1001$. The first case is finished.
The second case is more or less the same, except we are applying $3$ and $4$ to $5$. We must find the cases where the first case over counts multiples of five. Utilizing the floor function again on the fractions $\left\lfloor \frac{2001}{3\cdot5} \right\rfloor$, $\left\lfloor \frac{2001}{4\cdot5} \right\rfloor$, and $\left\lfloor \frac{2001}{3\cdot4\cdot5} \right\rfloor$ yields the numbers $133$, $100$, and $33$. The first two numbers counted all the numbers that were multiples of either four with five or three with five less than $2001$. The third counted the overlapping cases, which we must subtract from the sum of the first two. We do this to reach $200$. Subtracting this number from the original $1001$ numbers procures $\boxed{\textbf{(B)}\ 801}$.

Solution 3
First find the number of such integers between 1 and 2000 (inclusive) and then add one to this result because 2001 is a multiple of $3$.
There are $\frac45\cdot2000=1600$ numbers that are not multiples of $5$.  $\frac23\cdot\frac34\cdot1600=800$ are not multiples of $3$ or $4$, so $800$ numbers are.  $800+1=\boxed{\textbf{(B)}\ 801}$

Solution 4
Take a good-sized sample of consecutive integers; for example, the first $25$ positive integers. Determine that the numbers $3, 4, 6, 8, 9, 12, 16, 18, 21,$ and $24$ exhibit the properties given in the question. $25$ is a divisor of $2000$, so there are $\frac{10}{25}\cdot2000=800$ numbers satisfying the given conditions between $1$ and $2000$. Since $2001$ is a multiple of $3$, add $1$ to $800$ to get $800+1=\boxed{\textbf{(B)}\ 801}$.
~ mathmagical

Solution 5
By PIE, there are $1001$ numbers that are multiples of $3$ or $4$ and less than or equal to $2001$. $80\%$ of them will not be divisible by $5$, and by far the closest number to $80\%$ of $1001$ is $\boxed{\textbf{(B)}\ 801}$.
~ Fasolinka

Solution 5
Similar to some of the above solutions. 
We can divide $2001$ by $3$ and $4$ to find the number of integers divisible by $3$ and $4$. Hence, we find that there are $667$ numbers less than $2001$ that are divisible by $3$, and $500$ numbers that are divisible by $4$. However, we will need to subtract the number of multiples of $15$ from 667 and that of $20$ from $500$, since they're also divisible by 5 which we don't want. There are $133$ + $100$ = $233$ such numbers. Note that during this process, we've subtracted the multiples of $60$ twice because they're divisible by both $15$ and $20$, so we have to add $33$ back to the tally (there are $33$ multiples of $60$ that does not exceed $2001$). Lastly, we have to subtract multiples of both $3$ AND $4$ since we only want multiples of either $3$ or $4$. This is tantamount to subtracting the number of multiples of $12$. And there are $166$ such numbers. Let's now collect our numbers and compute the total: $667$ + $500$ - $133$ - $100$ + $33$ - $166$ = $\boxed{\textbf{(B)}\ 801}$.
~ PlainOldNumberTheory


Solution 6
Similar to @above:
Let the function $M_{2001}(n)$ return how many multiples of $n$ are there not exceeding $2001$. Then we have that the desired number is:
\[M_{2001}(3)+M_{2001}(4)-M_{2001}(3\cdot 4)-M_{2001}(3 \cdot 5) - M_{2001}(4 \cdot 5)+M_{2001}(3 \cdot 4 \cdot 5)\]
Evaluating each of these we get:
\[667+500-166-133-100+33 = 1100-299 = 801.\]
Thus, the answer is $\boxed{\textbf{(B)}\ 801}.$
-ConfidentKoala4