2001 AMC 10 Problem 25

First count the numbers up to $2001$ that are multiples of $3$ or $4$. Multiples of $3$: $\lfloor 2001/3 \rfloor = 667$. Multiples of $4$: $\lfloor 2001/4 \rfloor = 500$. Both (multiples of $12$): $\lfloor 2001/12 \rfloor = 166$. By inclusion-exclusion,
$$667 + 500 - 166 = 1001.$$

Now throw out those that are also multiples of $5$. A number divisible by $5$ and by $3$ or $4$ is a multiple of $15$ or of $20$. Multiples of $15$: $133$; multiples of $20$: $100$; multiples of both, i.e. of $60$: $33$. So
$$133 + 100 - 33 = 200$$
of the $1001$ numbers are divisible by $5$.

The count we want is $1001 - 200 = 801$.

The answer is $\boxed{\textbf{(B)}\ 801}$.

"$A$ or $B$ but not $C$" is the union of $A$ and $B$ with the part inside $C$ removed, and that removed part is itself a union: $(A \cap C) \cup (B \cap C)$. Each union is an inclusion-exclusion with overlaps measured by least common multiples. The number theory is routine; the whole difficulty is organizing the two-layer count without dropping an overlap.

The pattern repeats with period $60$. In $1$ to $60$ there are $20 + 15 - 5 = 30$ multiples of $3$ or $4$, of which $15, 20, 30, 40, 45, 60$ are multiples of $5$, leaving $24$. Since $2001 = 33 \cdot 60 + 21$, that gives $33 \cdot 24 = 792$ through $1980$, plus the qualifying numbers among $1981$ to $2001$ (nine of them: $1983, 1984, 1986, 1988, 1989, 1992, 1996, 1998, 2001$), for $801$.

• Subtracting all 400 multiples of 5 instead of only the multiples of 5 that are also multiples of 3 or 4.
• Forgetting the overlap term (multiples of $12$, or of $60$) in either inclusion-exclusion step, or computing $\lfloor 2001/12 \rfloor$ as $167$.