AMC12 2001 A

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 $10$, $15$, $20$, $30$, $40$, $45$ 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.