Difference between revisions of "2016 AMC 12A Problems/Problem 3"
(→Solution) |
(→Solution) |
||
| Line 1: | Line 1: | ||
| + | |||
| + | |||
==Solution== | ==Solution== | ||
<cmath>\text{rem}\left(\frac{3}{8},-\frac{2}{5}\right)</cmath> | <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> | + | <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)\lfloor -\frac{15}{16}\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}+\left(\frac{2}{5}\right)\left(-1\right)</cmath> | ||
<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> | ||
Revision as of 22:40, 3 February 2016
Solution
![\[\text{rem}\left(\frac{3}{8},-\frac{2}{5}\right)\]](http://latex.artofproblemsolving.com/4/b/c/4bc95cf8469a4cfdb6d3b62fdf68c7f9a237ba3b.png)
![\[=\frac{3}{8}-\left(-\frac{2}{5}\right)\Big\lfloor\frac{\frac{3}{8}}{-\frac{2}{5}}\Big\rfloor\]](http://latex.artofproblemsolving.com/7/1/9/7199ee4077d312c29bb69bff4eb734c8d19afbee.png)
![\[=\frac{3}{8}+\left(\frac{2}{5}\right)\Big\lfloor -\frac{15}{16}\Big\rfloor\]](http://latex.artofproblemsolving.com/5/f/8/5f8b7c703033544a74733c002937ae58f2e4eeaf.png)
![\[=\frac{3}{8}+\left(\frac{2}{5}\right)\left(-1\right)\]](http://latex.artofproblemsolving.com/c/e/1/ce191ca7c81c2c3d36c3c92798070445a4a58ca3.png)
![\[=\frac{3}{8}-\frac{2}{5}\]](http://latex.artofproblemsolving.com/0/e/8/0e82baa90861d265d469d327e817acadd6168a85.png)
![\[=\boxed{\textbf{(B)}-\frac{1}{40}}\]](http://latex.artofproblemsolving.com/f/f/e/ffe2006d31961de958874647aa4dcb4b583657fd.png)
