2001 AMC 10 Problem 8

1. Call today day $0$. Darren works on days $3, 6, 9, 12, \ldots$, the multiples of $3$. In the same way Wanda works on multiples of $4$, Beatrice on multiples of $6$, and Chi on multiples of $7$.
2. So the next day they are all together is the smallest number that is a multiple of $3$, $4$, $6$ and $7$ at the same time. This is called the least common multiple.
3. Break each number into primes: $3 = 3$, $4 = 2 \times 2$, $6 = 2 \times 3$, $7 = 7$.
4. The day must contain every one of those factors. It needs $2 \times 2$ (for $4$), a $3$ (for $3$ and $6$), and a $7$ (for $7$). The $6$ is already covered by the $2$ and the $3$.
5. So the day is $2 \times 2 \times 3 \times 7 = 84$.
6. Check: $84 \div 3 = 28$, $84 \div 4 = 21$, $84 \div 6 = 14$, $84 \div 7 = 12$. All whole numbers.

The answer is $\boxed{\textbf{(B)}\ 84}$.

"Every $n$th day" means the days that are multiples of $n$, so "everyone together again" means a common multiple, and "next" means the smallest one. The least common multiple is usually smaller than the product, because shared factors are only needed once; here the $6$ adds nothing new.

Check the choices from smallest to largest. $42$: $42 \div 4 = 10.5$, so Wanda is not there. $84$: divides evenly by $3$, $4$, $6$ and $7$. Since $84$ is the first choice that works, it is the answer.

Count days from today. Darren returns on days $3, 6, 9, \dots$ (multiples of $3$), Wanda on multiples of $4$, Beatrice on multiples of $6$, and Chi on multiples of $7$. All four are present exactly on the common multiples of $3, 4, 6, 7$, and the first such day is their least common multiple.

Factor: $3 = 3$, $4 = 2^2$, $6 = 2 \cdot 3$, $7 = 7$. The lcm takes the largest power of each prime:
$$2^2 \cdot 3 \cdot 7 = 84.$$

The answer is $\boxed{\textbf{(B)}\ 84}$.