2001 AMC 12 Problem 12

Multiples of $3$ or $4$ up to $2001$ number $$667 + 500 - 166 = 1001,$$ using $\lfloor \frac{2001}{3} \rfloor = 667,$ $\lfloor \frac{2001}{4} \rfloor = 500,$ and $\lfloor \frac{2001}{12} \rfloor = 166.$ Among these, the ones divisible by $5$ are multiples of $15$ or $20$: $$133 + 100 - 33 = 200,$$ using $\lfloor \frac{2001}{15} \rfloor = 133,$ $\lfloor \frac{2001}{20} \rfloor = 100,$ and $\lfloor \frac{2001}{60} \rfloor = 33.$ The count is $1001 - 200 = 801.$ Thus, the correct answer is B.