Logarithms

$\displaystyle \begin{aligned} \frac{6}{\log_{25} 360} + \frac{24}{\log_3 360} + \frac{9}{\log_{16} 360} & = \frac{6\log 25 + 24\log 3 + 9\log 16}{\log 360} \\ & = \frac{12\log 5 + 24\log 3 + 36\log 2}{\log 360} \\ & = 12 \ \frac{\log 5 + 2\log 3 + 3\log 2}{\log 360} \\ & = 12 \ \frac{\log 5 + \log 9 + \log 8}{\log (5 \cdot 9 \cdot 8)} \\ & = \boxed{12} \end{aligned}$

We use the fact that $\frac{1}{\log_{a}{b}} = \log_{b}{a}$ . So $\frac{6}{\log_{25}{360}}\ + \frac{24}{\log_{3}{360}}\ + \frac{9}{\log_{16}{360}} = \log_{360}{25^{6}} + \log_{360}{3^{24}} + \log_{360}{16^{9}} = \log_{2^{3}3^{2}5}{(5^{12}3^{24}2^{36})} = \boxed{12}$ .