1965 AHSME Problems/Problem 33

Revision as of 20:13, 13 February 2021 by Justinlee2017 (talk | contribs) (Created page with "==Solution== We can use Legendre's to find the number of <math>0</math>s in base <math>10</math> <cmath>\lfloor \frac{15}{5} \rfloor + \lfloor \frac{15}{25} \rfloor = 3</cmath...")
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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