2001 AMC 10 Problem 8

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 $\boxed{\textbf{(B)}\ 84}$.

"Every $n$-th day" is the set of multiples of $n$, and "the next day they are all back" is the smallest positive common multiple. Note that $6$ contributes nothing new once $2^2$ and $3$ are present; the lcm is not the product. Although dressed as a scheduling story, the whole problem is a single lcm computation.

Scan the choices from the smallest: $42$ is not a multiple of $4$; $84 = 4 \cdot 21 = 3 \cdot 28 = 6 \cdot 14 = 7 \cdot 12$ works for everyone. Since a smaller common multiple would have to divide $84$ and none of the choices below $84$ works, the answer is $84$.

• Multiplying 346*7 = 504 or forgetting that 4 requires a factor of 2^2, giving 42.
• Choosing $126 = 2 \cdot 3^2 \cdot 7$, which is not a multiple of $4$.