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

(Created page with "==Solution== <cmath>\text{rem}\left(\frac{3}{8},-\frac{2}{5}\right)</cmath> <cmath>=\frac{3}{8}-\left(-\frac{2}{5}\right)\lfloor\frac{\frac{3}{8}}{-\frac{-2}{5}}\rfloor</cmath...")
 
(Solution)
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<cmath>=\frac{3}{8}-\frac{2}{5}</cmath>
 
<cmath>=\frac{3}{8}-\frac{2}{5}</cmath>
 
<cmath>=\boxed{\textbf{(B)}-\frac{1}{40}}</cmath>
 
<cmath>=\boxed{\textbf{(B)}-\frac{1}{40}}</cmath>
 +
Notice that it would not matter if either <math>-\frac{2}{5}</math> or <math>\frac{2}{5}</math> were used.

Revision as of 22:15, 3 February 2016

Solution

\[\text{rem}\left(\frac{3}{8},-\frac{2}{5}\right)\] \[=\frac{3}{8}-\left(-\frac{2}{5}\right)\lfloor\frac{\frac{3}{8}}{-\frac{-2}{5}}\rfloor\] \[=\frac{3}{8}+\left(\frac{2}{5}\right)\lfloor -\frac{15}{16}\rfloor\] \[=\frac{3}{8}+\left(\frac{2}{5}\right)\left(-1\right)\] \[=\frac{3}{8}-\frac{2}{5}\] \[=\boxed{\textbf{(B)}-\frac{1}{40}}\] Notice that it would not matter if either $-\frac{2}{5}$ or $\frac{2}{5}$ were used.