Difference between revisions of "2016 AMC 12A Problems/Problem 3"

Latest revision as of 11:51, 4 February 2016

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+#REDIRECT [[2016 AMC 10A Problems/Problem 4]]
-==Solution==
-<cmath>\text{rem}\left(\frac{3}{8},-\frac{2}{5}\right)</cmath>
-<cmath>=\frac{3}{8}-\left(-\frac{2}{5}\right)\Big\lfloor\frac{\frac{3}{8}}{-\frac{2}{5}}\Big\rfloor</cmath>
-<cmath>=\frac{3}{8}+\left(\frac{2}{5}\right)\Big\lfloor -\frac{15}{16}\Big\rfloor</cmath>
-<cmath>=\frac{3}{8}+\left(\frac{2}{5}\right)\left(-1\right)</cmath>
-<cmath>=\frac{3}{8}-\frac{2}{5}</cmath>
-<cmath>=\boxed{\textbf{(B)}-\frac{1}{40}}</cmath>