The number of zeros to end a number is equal to the power of \(5\) in the \(n!.\) Since we are trying to find the lowest numbers, we can inspect small numbers where \(n \leq 60,\) so that the number of zeros to end \(n!\) is \[\left\lfloor \dfrac n5 \right\rfloor + \left\lfloor \dfrac n{25} \right\rfloor \] and the number of zeros to end \(2n!\) is \[\left\lfloor \dfrac {2n}5 \right\rfloor + \left\lfloor \dfrac {2n}{25} \right\rfloor .\] This means \[ 3\left(\left\lfloor \dfrac n5 \right\rfloor + \left\lfloor \dfrac n{25} \right\rfloor\right)\] \[= \left\lfloor \dfrac {2n}5 \right\rfloor + \left\lfloor \dfrac {2n}{25} \right\rfloor .\] Then, we can inspect each value of \(\left\lfloor \frac n5 \right\rfloor + \left\lfloor \frac n{25} \right\rfloor\) as follows: Case 1: If \( \left\lfloor \frac n5 \right\rfloor + \left\lfloor \frac n{25} \right\rfloor =0,\) then \(\left\lfloor \frac n5 \right\rfloor =0\) implying that \(0 \leq n < 4 \) which has no valid solutions. Case 2: If \( \left\lfloor \frac n5 \right\rfloor + \left\lfloor \frac n{25} \right\rfloor =1,\) then \(\left\lfloor \frac n5 \right\rfloor =1\) so \(5 \leq n < 10 .\) Then, we also know \[ \left\lfloor \dfrac {2n}5 \right\rfloor + \left\lfloor \dfrac {2n}{25} \right\rfloor =3.\] The only solutions are when \(\left\lfloor \frac {2n}5 \right\rfloor = 3,\) so \(15 \leq 2n < 20,\) making \(7.5 \leq n < 10.\) Thus, we have the solutions \(n = 8,9.\) Case 3: If \( \left\lfloor \frac n5 \right\rfloor + \left\lfloor \frac n{25} \right\rfloor =2,\) then \(\left\lfloor \frac n5 \right\rfloor =2\) so \(10 \leq n < 15 .\) Then, we also know \[ \left\lfloor \dfrac {2n}5 \right\rfloor + \left\lfloor \dfrac {2n}{25} \right\rfloor =6.\] The only solutions are when \(\left\lfloor \frac {2n}5 \right\rfloor = 5,\) so \(25 \leq 2n < 30,\) making \(12.5 \leq n < 15.\) Thus, we have the solutions \(n = 13,14.\) Since \(\left\lfloor \frac n5 \right\rfloor + \left\lfloor \frac n{25} \right\rfloor ,\) is non-decreasing, we found the four smallest solutions. As such, their sum is \(8+9+13+14=44,\) making the sum of the digits \(8.\) Thus, the correct answer is B.

2015 AMC 10B Exam Answer Key