2015 AMC 10B Problem 23

Factorials have far more factors of $2$ than $5$, so the number of trailing zeros of $m!$ is the number of factors of $5$:
$$Z(m) = \left\lfloor \frac{m}{5} \right\rfloor + \left\lfloor \frac{m}{25} \right\rfloor + \cdots$$
We need $Z(2n) = 3\,Z(n)$ with $n \ge 5$.

Roughly, $Z(2n) \approx 2\,Z(n)$, so tripling only happens when $(2n)!$ gains extra fives from multiples of $25$ that $n!$ lacks. For $n < 25$, $Z(n) = \lfloor n/5 \rfloor$, while $Z(2n) = \lfloor 2n/5 \rfloor + \lfloor 2n/25 \rfloor$. Tabulate by the block of $n$:

• $n = 5, 6, 7$: $Z(n) = 1$, $Z(2n) = Z(10), Z(12), Z(14) = 2$. No.
• $n = 8, 9$: $Z(n) = 1$, $Z(16) = Z(18) = 3$. Yes.
• $n = 10, 11, 12$: $Z(n) = 2$, $Z(20) = Z(22) = Z(24) = 4$. No.
• $n = 13, 14$: $Z(n) = 2$, $Z(26) = Z(28) = 6$. Yes.

Four values found: $8, 9, 13, 14$, and they are the least since we checked every $n \ge 5$ in order. Hence
$$s = 8 + 9 + 13 + 14 = 44,$$
whose digit sum is $4 + 4 = 8$.

The answer is $\boxed{\textbf{(B)}\ 8}$.

Trailing zeros are a question about the prime $5$ only, so Legendre's formula is the whole tool. The condition $Z(2n) = 3Z(n)$ is unusual because doubling $n$ normally doubles the fives; the slack comes from $25 \mid (2n)!$ before $25 \mid n!$, which localizes the solutions to $n$ just below multiples of $\frac{25}{2}$. A short table is faster than trying to solve the floor equation algebraically.

• Counting factors of 2 or of 10 instead of 5, or forgetting the extra contribution of 25 in (2n)! so that no n seems to work.
• Reporting $s = 44$ or the count of solutions instead of the digit sum, or including $n = 4$ or smaller despite the condition $n > 4$.