2013 AMC 10A Problem 17

Call yesterday day $0$; the period in question is days $1$ through $365$. Alice comes on multiples of $3$, Beatrix on multiples of $4$, Claire on multiples of $5$.

Two particular friends both come on multiples of the lcm of their periods:
- Alice and Beatrix: multiples of $12$, $\lfloor 365/12 \rfloor = 30$ days.
- Alice and Claire: multiples of $15$, $\lfloor 365/15 \rfloor = 24$ days.
- Beatrix and Claire: multiples of $20$, $\lfloor 365/20 \rfloor = 18$ days.

Each of these counts includes the days when all three come, the multiples of $\text{lcm}(3,4,5) = 60$: $\lfloor 365/60 \rfloor = 6$ days. Those must be removed from each pair count, because "exactly two" excludes them:
$$(30 - 6) + (24 - 6) + (18 - 6) = 24 + 18 + 12 = 54.$$

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

"Exactly two of three events" is (sum of pairwise overlaps) minus $3 \times$ (triple overlap): every triple-overlap day was counted once in each of the three pair counts. Periodic visits turn overlaps into lcm multiples, so the whole problem is three floor divisions plus one correction.

The pattern repeats every $60$ days. In days $1$ to $60$: multiples of $12$, $15$, $20$ number $5, 4, 3$, and only day $60$ has all three, giving $4 + 3 + 2 = 9$ exactly-two days per cycle. Six full cycles cover days $1$ to $360$: $54$. Days $361$ to $365$ contain no multiple of $12$, $15$ or $20$, so the total stays $54$.

• Counting multiples of 12, 15 and 20 without removing the multiples of 60, which gives 30 + 24 + 18 = 72.
• Subtracting the $6$ triple days only once instead of once from each pair, giving $66$.