1965 AHSME Problems/Problem 33

Solution

We can use Legendre's to find the number of $0$ s in base $10$ $\lfloor \frac{15}{5} \rfloor + \lfloor \frac{15}{25} \rfloor = 3$ So $h = 3$ . Likewise, we are looking for the number of $2^2$ s and $3$ s that divide $15!$ , so we use Legendre's again. $\lfloor \frac{15}{2} \rfloor + \lfloor \frac{15}{4} \rfloor + \lfloor \frac{15}{8} \rfloor = 7 + 3 + 1 = 11$ $\lfloor \frac{15}{3} \rfloor + \lfloor \frac{15}{9} \rfloor = 5 + 1 = 6$ Thus, $3^6 \vert 15!$ and $2^{11} \vert 15! \Rrightarrow (2^2)^5 \vert 15!$ So $k = 5$ , and $5+3 = 8$ $\boxed{D}$

~JustinLee2017

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1965 AHSME Problems/Problem 33

Solution