2001 AMC 10 Problem 25

Step 1: multiples of $3$ or $4$.

1. Multiples of $3$ up to $2001$: $2001 \div 3 = 667$, so there are $667$.
2. Multiples of $4$: $2001 \div 4 = 500.25$, so there are $500$.
3. Numbers that are multiples of both $3$ and $4$ are the multiples of $12$, and they were counted twice. $2001 \div 12 = 166.75$, so there are $166$.
4. Subtract the double count: $667 + 500 - 166 = 1001$ numbers are multiples of $3$ or $4$.

Step 2: remove the ones that are also multiples of $5$.

1. A number that is a multiple of $3$ and of $5$ is a multiple of $15$. A multiple of $4$ and of $5$ is a multiple of $20$. So we must remove the multiples of $15$ or $20$.
2. Multiples of $15$: $2001 \div 15 = 133.4$, so $133$. Multiples of $20$: $2001 \div 20 = 100.05$, so $100$.
3. Numbers that are multiples of both $15$ and $20$ are the multiples of $60$, counted twice. $2001 \div 60 = 33.35$, so $33$.
4. So $133 + 100 - 33 = 200$ of the $1001$ numbers are multiples of $5$.

Step 3: answer.

1. $1001 - 200 = 801$.

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

When counting "this or that," add the two counts and subtract the numbers counted twice. "But not $5$" means removing only the numbers already in the list that are also multiples of $5$, and that is another "this or that" count (multiples of $15$ or $20$). Keep the steps separate so no overlap is forgotten.

The pattern of multiples of $3$, $4$ and $5$ repeats every $60$ numbers. From $1$ to $60$, there are $20$ multiples of $3$ and $15$ of $4$, with $5$ counted twice, so $30$ are multiples of $3$ or $4$. Of these, $15, 20, 30, 40, 45, 60$ are multiples of $5$, leaving $24$. Since $33 \times 60 = 1980$, the numbers $1$ to $1980$ contain $33 \times 24 = 792$. The remaining numbers $1981$ to $2001$ that work are $1983, 1984, 1986, 1988, 1989, 1992, 1996, 1998, 2001$, which is $9$ more. Total: $792 + 9 = 801$.