2013 AMC 10A Problem 17

Call yesterday day $0$. The question is about days $1$ to $365$. Alice comes on days $3, 6, 9, \ldots$ (multiples of $3$), Beatrix on multiples of $4$, Claire on multiples of $5$.

1. When do Alice and Beatrix both come? On days that are multiples of both $3$ and $4$, which means multiples of $12$. From $1$ to $365$ there are $30$ of them, since $12 \times 30 = 360$.
2. Alice and Claire: multiples of $15$. There are $24$, since $15 \times 24 = 360$.
3. Beatrix and Claire: multiples of $20$. There are $18$, since $20 \times 18 = 360$.
4. When do all three come? On multiples of $3 \times 4 \times 5 = 60$. There are $6$, since $60 \times 6 = 360$.
5. Take those out. A day when all three come is not an "exactly two" day. Each such day was counted in all three lists above (for example, day $60$ is a multiple of $12$, of $15$ and of $20$). So remove $6$ from each list:
$$(30 - 6) + (24 - 6) + (18 - 6) = 24 + 18 + 12 = 54.$$

The answer is $\boxed{\textbf{(B)}\ 54}$.

Two repeating schedules line up on multiples of the least common multiple of their gaps. "Exactly two" means you must throw out the days when the third friend also shows up. Each all-three day sits in every one of the three pair lists, so it is removed three times, once from each.

The whole pattern repeats every $60$ days, so look at days $1$ to $60$ first. Multiples of $12$: $5$ days. Of $15$: $4$ days. Of $20$: $3$ days. Only day $60$ has all three, so exactly-two days number $4 + 3 + 2 = 9$ per block of $60$. Six blocks cover days $1$ to $360$, giving $6 \times 9 = 54$. Days $361$ to $365$ contain no multiple of $12$, $15$ or $20$, so the answer stays $54$.